Bring it up using the method of mathematical induction for anyone. Science's start

Savelyeva Katerina

The activity is considered to be based on the method of mathematical induction at the highest level of divisibility, before adding series. The application of the method of mathematical induction to the proof of irregularities and the highest geometrical tasks is examined. The work is illustrated with a presentation.

Vantage:

Forward view:

Ministry of Science and Development of the Russian Federation

Sovereign pledge

secondary school no. 618

Course: algebra and basic analysis

Theme of project work

“Method of mathematical induction and stagnation to the highest specification”

Viconal robot: Savelyeva E, 11B class.

Kerivnyk : Makarova T.P., teacher of mathematics, GOU ZOSH No. 618

1. Introduction.

2. The method of mathematical induction in the highest task for completeness.

3. Use of the method of mathematical induction before summing up the series.

4.Apply the method of mathematical induction until the irregularities are confirmed.

5. Use of the method of mathematical induction until the geometric tasks are resolved.

6. List of Wikipedia literature.

Enter

The basis of any mathematical investigation is deductive and inductive methods. The deductive method of merchandising is the merkuvannya from the hidden to the private, then. fading, the final moment of which is a hidden result, and the final moment is a private result. The induction stagnates during the transition from private results to secret ones, then. є by a method ranging from deductive. The method of mathematical induction can be equalized with progress. We start from the bottom, as a result logical idea We're getting to the point. The people have already given up progress, to develop their thoughts logically, since nature itself has designed it to grow inductively. Although the scope of the method of mathematical induction has grown, the school curriculum devotes little time to it. But it’s so important – consider the dimensions inductively. The established principle is now in high demand and the proof of the theorems is in line with the study of school practice and other mathematical principles: the inclusion of the third, the inclusion-exclusion, Dirichlet and others. This essay contains knowledge from various branches of mathematics, the main tool of which is the use of the method of mathematical induction. Speaking about the importance of this, A.N. Kolmogorov noted that “reason and understanding establish the principle of mathematical induction as a good criterion of maturity, which is necessary for mathematics.” The method of induction in its entirety lies in the transition from private security to a universal one, legal regularities and legal formulations. This research method is, of course, the main method of conducting research in any experimental natural science.

human activity. The method (principle) of mathematical induction in its simplest form comes to a standstill when it is necessary to complete the derivation of all natural numbers.

Zavdannya 1. In the article “How I became a mathematician” by A.N. Kolmogorov writes: “I recognized the joy of mathematical “discovery” early, having noticed a pattern in the five or six rocks

1 =1 2 ,

1 + 3 = 2 2 ,

1 + 3 + 5 = 3 2

1 + 3 + 5 + 7 = 4 2 and so on.

At school I saw the magazine "Vesnyany Lastivki". No one has published my opinion...”

We don’t know what kind of evidence there is of this journal’s guidance, until everything began under private security. The hypothesis itself, which, melodiously, disappeared after the discovery of these private jealousies, suggests that the formula

1 + 3 + 5 + ... + (2n - 1) = n 2

true for any given number n = 1, 2, 3, ...

To confirm this hypothesis, it is enough to establish two facts. First of all, for n = 1 (i for n = 2, 3, 4) it is necessary to affirm correctly. In another way, it is acceptable that the affirmation is more correct when p = before, and is reconverted, which is also true for n = to + 1:

1 + 3 + 5 + ... + (2k - 1) + (2k + 1) = (1 + 3 + 5 + ... + (2k - 1)) + (2k + 1) = up to 2 + ( 2k + 1) = (K + I) 2.

Well, the firmament, what is brought about, is correct for everyone n: for n = 1 is true (not verified), and from another fact - for n = 2, stars for n = 3 (through the same, different fact) then.

Lesson 2. Let's take a look at all possible primary fractions with number 1 and whatever (for the purpose of putting-

him) znamennik: Bring that for whatever p> 3 you can see one at a glance P different types of shot.

Rishennya, Let's recheck the hardness when n = 3; maєmo:

Otzhe, basic firmament viconano

Let us now accept that the firmament that we will be honored is true for any number before, and we will prove that it is true for the following date before + 1. In other words, it is acceptable that the manifestation

in yakoma k Dodanki and all the banners of the massacre. We will show that you can also remove the given unit from the view of the sum before + 1 fractions of the required appearance. It is important to note that when the fractions change, then the banners (at the given one with a sum before Dodankov) grow evil to the right so that T - The greatest of the famous people. We are denying our need for tribute from the view of the sum(before + 1)th fraction, when one fraction is divided into two, for example the remaining one. You can earn some scraps

And that

In addition, all the fractions were lost, fragments T was the greatest banner, and t + 1 > t, i

t(t+1) > t.

Thus, we have established:

1. with n = 3 tse firmness is correct;

1. What is right for us is true for us before,
   then it is more correct for up to +1.

On this basis we can confirm that the statement we see is correct for all natural numbers, starting with three. Based on the evidence generated, an algorithm is generated to find the necessary distribution of the unit. (What kind of algorithm is this? Submit the number 1 for the sum of 4, 5, 7 dodanki independently.)

With the top of the front two, the order was divided into two pieces. The first croc is called basis induction, other -inductive transitionwith a short period of induction. The other term is the most important, and it includes submissiveness (the assertion is true when n = k) i conjecture (confirmation is true when n = to + 1). The parameter itself is called induction parameterThis is a logical scheme (technique), which allows for structure, so that if you look at the statements of the correct natural numbers (and all, starting from this), the parts of the validity of both the basis and the transition are calledthe principle of mathematical induction, in any way The method of mathematical induction was founded.The term “induction” itself is similar to the Latin word induction (guidance), which means the transition from single knowledge about various objects of this class to general knowledge about all objects of this class, which is one of the main methods of knowledge.

The principle of mathematical induction, itself in the basic form of two terms, first appeared in 1654 in Blaise Pascal’s work “Treatise on the Arithmetic Tricet”, in which induction introduced a simple way of calculating the number nominal coefficients). D. Polya at the bottom quotes B. Pascal with small changes that are given to square arms:

“Regardless of those who consider the proposition [obviously a formula for binomial coefficients] to avenge the blindness of private attacks, I will give a very short proof for it, based on two points.

The first lemma confirms that stewing is correct for falling asleep - this is obvious. [Pri P = 1 the formula is explicit...]

Another lemma confirms the approach: if our assumption is true for a sufficient basis [for a sufficient result], then it will be true for the basis following it [for n+1].

Between these two, the justice of the saying for all meanings inevitably shines through. P. True, from the first lemi it is true for P = 1; However, due to another principle, it is fair for P = 2; Well, I know, by virtue of another principle, it’s fair for n = 3 and so on ad infinitum.”

Challenge 3. The “Veighes of Hanoi” puzzle consists of three puzzles. On one of the strands there is a pyramid (Fig. 1), which consists of many rings of different diameters, which change from bottom to top

Fig 1

This pyramid needs to be moved to one of the other strands, transferring one ring at a time and not placing more rings on the smaller one. How can you earn money?

Decision. Well, we need to respond to the power supply: how can we move the pyramid that is formed by P a ring of different diameter, from one cut to another, following the rules of the cut? Now the data we have, as it seems, is parameterized (a natural number has been introduced d), This can be determined by the method of mathematical induction.

1. The basis of induction. When n = 1 everything is clear, because the pyramid from one ring can obviously be moved to any kind of haircut.
2. Induction time. It is acceptable that we can move any kind of pyramids around a number of rings p = up to.
   Let's see that we can also move the food from n = up to +1.

Pyramidka from to ring, what to lie on the greatest(before + 1) rings, we can, with allowances, move them to any other cut. Zrobimo tse. Neruhome(before + 1)th ring we will not be required to carry out the relocation algorithm, as much as possible. After relocation before ring, moveable to the greatest extent(before + 1)th ring for a haircut that was lost. And then the displacement algorithm known to us behind inductive assumptions again stagnates before ring, and move them to the cutting from those lying at the bottom(before + 1)th ring. In such a manner that we move the pyramids from before rings, then instead of moving the pyramids from before + 1 rings. Then, based on the principle of mathematical induction, you can now move the pyramid that adds up to p kіlets, de p > 1.

The method of mathematical induction in the highest task for completeness.

Using the additional method of mathematical induction, it is possible to prove various assertions about the divisibility of natural numbers.

Zavdannya 4 . If n is a natural number, then the number is par.

When n=1 our statement is true: the guy is number. Let's say it's a guy's number. Oskolki, a 2k is a guy’s number, that’s a guy’s number. Also, the parity is shown at n=1, and from the parity the parity is inferred. Well, in pairs for all natural values ​​of n.

Zavdannya 3. Bring the number Z 3 + 3 - 26n - 27 with sufficient natural n divided by 26 2 without excess.

Decision. We will advance by induction an additional solidification, which is 3 3n+3 - 1 is divided by 26 without excess when n>0.

1. The basis of induction. For p = 0 maєmo: 3 3 - 1 = 26 -divided by 26.

Induction time. Let's say 3 3n+3 - 1 is divided by 26 when p = before, ta Let us know that in this case the affirmation will be true p = up to + 1. Shards 3

then from the inductive assumption we can conclude that the number is 3 3k + 6 – 1 is divided by 26.

Now the firmament has been brought to light, formulated in the minds of the people. I will return with induction.

1. The basis of induction. Obviously, when n = 1 stronghold: fragments 3 3+3 - 26 - 27 = 676 = 26 2 .
2. Induction time. It is acceptable that when n = up
   viraz 3 3k + 3 - 26k - 27 is divided by 26 2 without further ado, and let us prove that the affirmation is correct when n = to + 1,
   what is the number

divide by 26 2 no extra charge. In the remaining amount, the resentment of the dodanks will be divided without excess by 26 2 . Pershe - to the one who brought the perfect look to stand at the arms, on 26; another - for the administration of induction. Based on the principle of mathematical induction, the necessary assertion has been fully demonstrated.

Use of the method of mathematical induction before summing up the series.

Zavdannya 5. Finish the formula

N is a natural number.

Decision.

When n = 1, the offending parts of jealousy are converted to one and, therefore, first of all the principle of mathematical induction.

Let's say the formula is correct for n=k, then.

It is possible to add to both parts this equality and reconcile the right part. Todi is removable

Thus, since the formula is true for n=k, it follows that it is true for n=k+1. This statement is fair for any natural value k. Also, a friend of the principle of mathematical induction is also Vikonian. The formula has been completed.

Zavdannya 6. There are two numbers written on the doshtsa: 1,1. Having entered the sum between the numbers, we subtract the numbers 1, 2, 1. By repeating this operation one more time, we subtract the numbers 1, 3, 2, 3, 1. After three operations there will be numbers 1, 4, 3, 5, 2, 5, 3, 4, 1. What will be the sum of all numbers for the next day? 100 operations?

Decision. Vikonuvati usi 100 The operations would be even laborious and distressing. Well, you need to try to know the secret formula for sumi S numbers after p operations. Let's look at the table:

Have you noticed any pattern here? However, you can earn one more coin: after several operations there will be numbers

1, 5, 4, 7, 3, 8, 5, 7, 2, 7, 5, 8, 3, 7, 4, 5, 1,

the sum of which S 4 is equal to 82.

In fact, you can not write down the numbers, but immediately say how the amount will change after adding new numbers. Let the sum rise 5. What will you be like if new numbers arrive? There is a new number of skins for the sum of two old ones. For example, from 1, 3, 2, 3, 1 we go to 1,

1 + 3, 3, 3 + 2, 2, 2 + 3, 3, 3 + 1, 1.

If the same number (besides the two extreme ones) is now included in the sum three times, then the new sum is equal to 3S - 2 (2 is added to add the ones that are daily). Tom S 5 = 3S 4 - 2 = 244, i vzagali

What kind of secret formula is that? If it were not just two alone, then the sum would immediately increase in the morning, as at the steps of the three (1, 3, 9, 27, 81, 243, ...). And our numbers, apparently, are one more. In this manner, you can let go

Let’s now try to complete this by induction.

The basis of induction. Marvel at the table (for n = 0, 1, 2, 3).

Induction time. Let's say

Let's bring it up then S to + 1 = 3 to + 1 + 1.

True,

Well, our formula has been completed. It can be seen that after a hundred operations the sum of all numbers on the modern day 100 + 1.

Let's look at one wonderful example of the principle of mathematical induction, in which you first need to set two natural parameters and then carry out induction using them.

Zavdannya 7. Bring what you want= 2, x 2 = 3 and for whatever natural p> May 3rd is the place of marriage

x p = 3x p - 1 - 2x p - 2

That

2 p - 1 + 1, p = 1, 2, 3, ...

Decision. Dear, whose given output is the sequence of numbers(x p) is indicated by induction, the remaining members of our sequence, except the first two, are specified inductively, then through the front ones. This is what the sequence of tasks is called recurrent, And in our view, this sequence is designated (to the duties of the first two members) by a single rank.

The basis of induction. The result consists of reversing two pillars: p = 1 i p = 2. In both cases, the firmness is fair behind the mind.

Induction time. Let's say what for p = up to - 1 i p = up to viconano stronghold, tobto

Let us then bring justice to the confirmation for n = to + 1. Mayo:

x 1 = 3 (2 + 1) - 2 (2 + 1) = 2 +1, which is what needs to be completed.

Zavdannya 8. Show that, even if it is a natural number, it can be represented by the sum of many different terms of the recurrent sequence of Fibonacci numbers:

at up to > 2.

Decision. Let's go - natural number. We will carry out induction as soon as possible P.

The basis of induction. When n = 1 affirmation is fair, since 1 itself is a Fibonacci number.

Induction time. It is acceptable that all natural numbers are smaller in number P, You can apply the sum of several different members of the Fibonacci sequence. We know Fibonacci best Ft, I don’t overdo it P; in this order, F t p i F t +1 > p.

Oskolki

After the induction is allowed, the number p-F t can be presented in the form of 5 different members of the Fibonacci sequence, and with the remaining inequality, all members of the Fibonacci sequence take part in the sum and 8, less Ft. So the numbers are laid out n = 8 + F t satisfies the mind with excitement.

Apply the method of mathematical induction until the irregularities are confirmed.

Zavdannya 9. (Bernoulli's anxiety.)Let me know what x > -1, x 0, і zhalom n > 2 unfairness is fair

(1+x) p>1+xp.

Decision. The proof is again carried out by induction.

1. The basis of induction. We are transforming justice into inequality when n = 2. True,

(1 + x) 2 = 1 + 2x + x 2> 1 + 2x.

2. Induction period. Acceptable, what for the number n = up the firmament is just, then

(1 + x) to > 1 + xk,

De to > 2. We bring yogo for n = to + 1. Maєmo: (1 + x) to + 1 = (1 + x) to (1 + x)> (1 + x) (1 + x) =

1+(k+1)x+kx 2 > 1+(k+1)x.

However, on the basis of the principle of mathematical induction, it can be confirmed that Bernoulli’s inequality is valid for any n>2.

Not always in the minds of the task that lies behind this method of mathematical induction, there is a clearly formulated hidden law that follows. Sometimes, one has to be careful to identify (guess) the severe fallouts of the kidneys, to establish any fundamental law of the stink, and then to arrive at the established hypothesis using the method of mathematical induction. In addition, a significant induction may be masked, and first of all, it is necessary to determine what parameter the induction is based on. Let's look at this misfortune anyway.

Zavdannya 10. Bring it out

whatever the natural n>1.

Rishennya, Let's try to bring up this inequality using the method of mathematical induction.

The induction basis can be easily verified:1+

After induction

and we need not convey that

How to speed up inductive assumptions, we confirm that

Although jealousy is truly true, it does not give us the greatest happiness.

Let's try to make it stronger than necessary for the exit task. And let us tell you that

You may think that bringing the goal of induction to the right is hopeless.

However, for n = 1 maєmo: the firm is correct. For priming the inductive coating, it is acceptable that

And let us tell you that

True,

In this manner, we have conveyed a stronger affirmation, from which the affirmation immediately emerges, which takes place in the mind of the faithful.

The main ones here are those that we wanted and had to make stronger assertions, without requiring the given ones, otherwise we could quickly end up with strong allowances in the inductive process. This explains that the straightforward principle of mathematical induction leads to the conclusion.

The situation that arose at the hour of great tragedy was given a namewinemaker phenomenon.The phenomenon itself lies in the fact that more complex plans can be implemented with great success, since they are based on a greater depth of intelligence.

Command 11. Bring that 2 t + p – 2 tp for whatever natural people type.

Decision. There are two parameters here. So you can try to hold such a callsecondary induction(Induction in the middle of induction).

We will carry out inductive merging thoroughly P.

1. Induction base per p. When n = 1 you need to check what 2t~1>t. To prove this inequality, rapid induction of water T.

A) The basis of induction for the like. When t = 1 sign
jealousy, which is permissible.

b) Induction period for etc.It is acceptable that when t = to the firmament is true, then 2 to ~ 1 > to. Todi until
It seems that the affirmation will be true even when t = up to +1.
Maemo:

with natural up to.

In this manner, nervousness 2 stick to whatever is natural T.

2. Induction time per p.Selectable and fixed as a natural number T. It is acceptable that when n = I the firmness is fair (with a fixed t), then 2 t +1 ~ 2 > t1, and we will prove that this affirmation will be fair and n = l+1.
Maemo:

for whatever natural people t ta p.

Also, on the basis of the principle of mathematical induction (for d) solidification of the commandment is correct for whatever P and for any fixed T. In this way, this inequity comes down to any natural type.

Zavdannya 12. Let's go - natural numbers, and t > p. There are two more numbers:

Skin infections before square root signs, t i p are afraid.

Decision. Let's bring this to the next point.

Lemma. For whatever natural people t i p (t > p) and to the invisible (not necessarily to the whole) X unfairness is fair

Finished. Let's take a look at the nervousness

This inequality is fair, because the resentment of peers on the left side is positive. Opening the arches and re-creating them, we remove them:

The elastic square root from both parts of the remaining unevenness is removed by hardening with lema. Well, the problem has been reached.

Let us now move on to the solution to the problem. Significantly higher than the given numbers through A, and to the other - through b to. Let's see what whatever the natural before. The proof is carried out using the method of mathematical induction, strictly for the guys and the non-pairs before.

The basis of induction. When up = 1 may be uneasy

y[t > y/n , fair through those who t > p. When up to = 2 it is necessary to leave with the substitution given by Lema x = 0.

Induction time. Acceptable, in case of actual to nervousness a >b to fair. Let's see what

With the assumption of induction and the monotonicity of the square root, we can:

On the other side, on the other side, the lema is pouring,

Combining two remaining inequalities, we can eliminate:

This conclusion has been brought to the principle of mathematical induction.

Zavdannya 13. (Koshy's nervousness.)Let us know that for any positive numbers..., a p unfairness is fair

Decision. When n = 2 nervousness

We are very aware of the arithmetic mean and the geometric mean (for two numbers). Let's go n = 2, up to = 1, 2, 3, ... and we will now carry out induction on before. The base of the induction value is in place Having assumed now that the necessary inequality is already installed for n = 2, let's bring it to you for P = 2. Maemo (stagnant inequality for two numbers):

Well, for the induction indulgences

In this way, by induction with us, we created inequality for everyone p 9 - step of two.

To prove inequality for other values P By accelerating “downward induction”, we can see that the unevenness of Viconan is for the rather unknown P numbers, then the same is true for(P - 1st day. Let's get over it, respectfully, for the sake of the zroblenim allowances for P numbers Wiconano unevenness

then a g + a 2 + ... + a n_x> (n - 1) A. Having divided the offending parts into P - 1, the necessary nervousness is eliminated.

Now, we have established from the beginning that inequality has a place for an infinite number of possible values. P, and then they showed that the unevenness of Wiconan is for P numbers, then this is true for(P - 1) numbers. The situation is now settled, so that the cat’s uneasiness may be the place for a set of P be-any unknown numbers for be-something n = 2, 3, 4, ...

Zavdannya 14. (D. Uspensky.) For any trikutnik ABC, for any kuti = CAB, = CBA dusk, a place of uncertainty looms

Decision. Kuti and twilight, ale tse (behind the meanings) means that tse kuti are in the dark world, as = p, = (p, q are mutually prime natural numbers).

A quick method of mathematical induction is carried out using the bag p = p + q mutually prime natural numbers.

The basis of induction. When p + q = 2 we can: p = 1 and q = 1. Then the tricutus ABC is equal, and the necessary inequalities are obvious: stinks emanate from the tricutus imbalance

Induction time. It is now acceptable that the inequalities are set for p + q = 2, 3, ..., to - 1, de to > 2. Let us prove that inequalities are fair p + q = k.

Let's go ABC - Danish trikutnik, in which> 2. Both sides AC and PS can't be jealous: don't let me AC > ND. Now let's talk about baby 2, the equilateral tricubitus ABC; maєmo:

AC = DC i AD = AB + ВD, then,

2AC > AB + BD (1)

Let's take a look now at the trikutnik GVA, Anything can also be equalized:

DСВ = (q - р), ВDC = p.

Small 2

For which knitted sweater is the inductive suspense given, and to that

(2)

Adding (1) and (2), we can:

2AC+BD>

and that

From the same trikutnik VBS after the induction is started, it is necessary to

Doctor's anterior unevenness, we lay it down

Thus, the inductive transition is eliminated, and the hardening of the problem follows from the principle of mathematical induction.

Respect. The firmness of the promise is lost in strength when the situation is dull. At the core of this approach, we now have to establish another important mathematical principle - the principle of non-interruption.

Order 15. Straight pieces divide the surface into pieces. Let us know that these parts can be prepared from whites

and black color so that the vessel parts that form the cordon cordon are in a different color (like for baby 3 n = 4).

pic 3

Decision. Speedy induction over a number of straight lines. Oh dear, let me go P - The number of straight lines that divide our area into parts, n>1.

The basis of induction. I'm just alone(P = 1), then you can divide the surface into two surfaces, one of which can be prepared into white color, and the other one is black, and the confirmation is true.

Induction time. To prove the inductive transition more clearly, let’s look at the process of adding one new direct line. Let's tell a friend directly(P= 2), then we can remove four parts that can be prepared in a proper manner, having prepared the beds of the same color. I’ll wonder what will happen if we carry out the third straight line. You can divide the sections of the “old” parts to create new sections of the cordon, with the same colors on both sides (Fig. 4).

Small 4

Let's put it this way:from one sidein the new direct we change the colors - white and black; In this case, those parts that lie on the other side in a straight line are not over-prepared (Fig. 5). Then this new painting will satisfy the required requirements: on one side the straight line was already drawn (along with other colors), and on the other side it was required. In order for the parts that form the cordon to be held in a straight line, they are prepared in different colors, and the parts are replenished only on one side in front of the straight line.

Fig.5

We will now complete the inductive transition. Acceptable, what for the actionn = upthe confirmation of the legacy is fair, so that all parts of the area on which they will sharebeforeDirectly, you can prepare it in white and black colors so that the bottom parts are of a different color. Let us see that such a barrage is also forP= before+ 1 Straight It is similar to the transition from two straight lines to three. We will hold it on the squarebeforestraight Then, after the induction has been started, the removed card can be properly processed. Let's do it now(before+ 1)-th straight line and on one side in front of it we change the colors on the bed. In this manner, now(before+ 1) right here the plots are divided into carved colors, with which the “old” parts, which we have already prepared, are no longer properly prepared. Similar to the principle of mathematical induction.

Zavdannya16. On the edge of the desert there is a large supply of gasoline and a car that can travel 50 kilometers with repeated refills. There are also a number of canisters in which you can pour gasoline from the car’s gas tank and store it for saving at any empty point. Let us know that the car can travel for more than 50 kilometers, regardless of whether it’s complete. You are not allowed to carry gasoline cans; you can carry empty ones at any location.

Decision.Let's try to bring the induction to speedP,what the car can be driven onPkilometers from the edge of the desert. AtP= 50 vidomo. It was impossible to carry out the induction procedure and explain how to get theren = up+ 1 kilometers, as you known = upYou can travel kilometers.

However, here we find it difficult to understand: after we passedbeforekilometers, gasoline may not last on the way back (not to mention saving money). And in this situation, the way out lies in the strengthening of the firmness that is being achieved (the winemaker’s paradox). We'll show you that you can just get byPkilometers, and also build up a large supply of gasoline at the same time at the stationPkilometers from the edge of the desert, stopping at this point after completing the transportation.

The basis of induction.Let alone gasoline - the amount of gasoline is necessary to make one kilometer of road. If a flight goes 1 kilometer and back, it takes two units of gasoline, then we can save 48 units of gasoline at the same distance of a kilometer from the edge and turn back for a new portion. In this way, for a few trips before the meeting, you can build up a supply of sufficient size that will be needed. If you need to add 48 units to your reserve, you will spend 50 units on gasoline.

Induction time.It’s acceptable that on the roadP= beforeAt the edge of the desert you can stock up on some gasoline. Let us prove that it is possible to create a contusion on the risen = up+ 1 kilometer with a predetermined supply of gasoline and check with such a vehicle for transportation. Shards exactlyP= beforeIf there is no supply of gasoline, then (based on the induction base) we can make several trips to the pointn = up+ 1 earn in pointP= before4- 1 stock of sufficient size as needed.

The truth of the forbidden assertion, which was not previously known, now emerges from the principle of mathematical induction.

Visnovok

Since then, having learned the method of mathematical induction, I have advanced my knowledge of this field of mathematics, and also began to overcome knowledge that was previously beyond my power.

This has logical and purposeful tasks, then. the very ones that advance interest to mathematics itself as a science. The majority of such tasks become useful activities and can lead to new additions to mathematical labyrinths. In my opinion, this is the basis of any science.

By continuing to learn the method of mathematical induction, I will learn to understand this not only in mathematics, but also in the most important problems of physics, chemistry and life itself.

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63s.

11.Solominsky I.S., Golovina L.I., Yaglom I.M. About mathematical induction. - M: Science. – 1967. – P.7-59.

12.http://w.wikimedia.org/wiki

13.htt12://www.refeshtcollestiop.ru/40 124.html

p align="justify"> The method of proof, based on the Peano axiom 4, is used to prove many mathematical powers and various principles. The basis for this is the following theorem.

Theorem. What is the firmness A(n) with natural meat n true for n= 1 from what is true for n = k trace that it is true and for the date n=k, those strongholds A(n) n.

Finished. Significantly through M without these and without these natural numbers, for some solids A(n) true. Here's the theorem: 1) 1 M; 2) kMkM. Stars, on the stand of axiom 4, we place what M =N, then. hardening A(n) true for any natural n.

The method of proof that is based on this theorem is called using the method of mathematical induction, and the axiom is the axiom of induction. Such proof consists of two parts:

1) bring it to hardening A(n) true for n= A(1);

2) allow it to remain firm A(n) true for n = k, and, emerging from this suspension, bring about what is firmly established A(n) true for n = k + 1, then. What is true love A(k) A(k + 1).

Yakshcho A( 1) A (k) A(k + 1) - correct vislovlyuvannya, then to break the visnovok from the one who is firmly A(n) true for any natural number n.

The proof by the method of mathematical induction can begin not only by confirming the truth of the statement for n= 1, if any natural number m. Who's got it hardened A(n) will be shown for all natural numbers nm.

Zavdannya. Let us prove that for any natural number the true equal is 1 + 3 + 5 … + (2 n- 1) = n.

Decision. Jealousy 1 + 3 + 5 … + (2 n- 1) = n- a formula that can be used to find the sum of the first consecutive unpaired natural numbers. For example, 1 + 3 + 5 + 7 = 4 = 16 (amount of 4 additional danks), 1 + 3 + 5 + 7 + 9 + 11 = 6 = 36 (sum of 6 additional danks); If this sum is enough to place 20 additions to the designated type, then it is equal to 20 = 400, etc. Having established the truth of this equation, it is possible to find from the formula the sum of any number of additions of the designated type.

1) Converted to the truth of this equality for n= 1. When n= The 1 left side of equalness is composed of one term equal to 1, the right side is equal to 1 = 1. So since 1 = 1, then for n= 1 ceremonial is true.

2) It is acceptable that this jealousy is true for n = k, then. what 1 + 3 + 5 + … + (2 k- 1) = k. Coming from this assumption, it is clear that this is true for n = k + 1, then. 1 + 3 + 5 + … + (2 k- 1) + (2(k + 1) - 1) = (k + 1).

Let's look at the left side of the remaining zeal.

For the indulgences, the scrip of the first k Dodankov is ancient k and volume 1 + 3 + 5 + … + (2 k- 1) + (2(k + 1) - 1) = 1 + 3 + 5 + … + (2k- 1) + (2k+ 1)=

= k+(2k + 1) = k+ 2k + 1. Viraz k+ 2k + 1 equally ancient virus ( k + 1).

Well, the truth of this equality for n = k + 1 completed.

In this manner, true zeal is given to n= 1 and from the truth of it for n = k trace truth for n = k + 1.

Tim himself realized that this jealousy is true for any natural number.

Using a similar method of mathematical induction, it is possible to establish the truth of both equities and inequalities.

Zavdannya. Bring it on nN.

Decision. Let's verify the truth of inequality when n= 1. Maemo – real anxiety.

It is acceptable that inequality is true when n = k, tobto. - Helpful anxiety. Let us prove, without prejudice, that it is more correct and when n = k + 1, then. (*).

Let’s resolve the left part of the nervousness (*), with medical help: .

Ale, that means i .

Well, this inequality is true for n= 1, and from the fact that inequality is true for the active n= k, we rejected what was more correct for n= k + 1.

Tim himself, vikorista and axiom 4, brought us to the conclusion that this inequality is true for any natural number.

Using the method of mathematical induction, other propositions can be developed.

Zavdannya. Bring that any natural number is a true solid.

Decision. Let's verify the truth of the confirmation when n= 1: -Registration.

It is acceptable which is more correct when n = k: . Let us show, vikorista, the truth of confirmation in n = k + 1: .

Convertible Viraz: . Let's know the sacristy kі k+ 1 members. It appears that the difference is a multiple of 7, and after the omissions it appears to be divisible by 7, then the multiple is also changed to 7:

Additions are divisible by 7, then i.

In this manner, the principle is true for n= 1 and from the truth of it for n = k trace truth for n = k + 1.

Tim himself realized that this affirmation is true for any natural number.

Zavdannya. Bring what for any natural number n 2 true affirmations (7-1)24.

Decision. 1) We verify the truth of the confirmation when n= 2: - Helpful statement.

Mathematical induction is the basis of one of the most extensive methods of mathematical proof. With this help you can complete most of the formulas with natural numbers n, for example, the formula for finding the sum of the first terms of progress S n = 2 a 1 + n - 1 d 2 · n Newton's binomial formula a + b n = C n 0 · a n · C n 1 · a n - 1 · b +. . . + C n n - 1 · a · b n - 1 + C n n · b n .

At the first point, we will look at the basic concepts, then we will look at the foundations of the method itself, and then we will understand how jealousy and inequality can be achieved.

Concepts of induction and deduction

Let’s now look at what induction and deduction are all about.

Viznachennya 1

Induction- this is a transition from private to private, and deduction navpaki – from zagalnogo to chastkovy.

For example, we have a saying: 254 can be divided into two parts. From this we can develop a wide range of weapons, some of which will be useful and some of which will be harmful. For example, the assertion that all whole numbers, such as the number 4, can be divisible by two without excess - true, and those that are a number of three signs can be divisible by 2 - hibna.

In general, we can say that with the help of inductive annihilation, it is possible to remove the absence of substances from one visible and obvious oxidation. Mathematical induction allows us to determine how fair the results are.

Let us assume that we have a sequence of numbers like 1 1 · 2 , 1 2 · 3 , 1 3 · 4 , 1 4 · 5 , . . . , 1 n (n + 1), where n means a natural number. In this case, when the first elements are folded, the sequence we take away is:

S 1 = 1 1 2 = 1 2, S 2 = 1 1 2 + 1 2 3 = 2 3, S 3 = 1 1 2 + 1 2 3 + 1 3 4 = 3 4, S 4 = 1 1 · 2 + 1 2 · 3 + 1 3 · 4 + 1 4 · 5 = 4 5 , . . .

By vicoristic induction, one can proceed unpretentiously, so that S n = n n + 1 . In the third part we will complete this formula.

What is the method of mathematical induction?

It is based on the same principle. It is formulated as follows:

Vicennia 2

If this assertion is valid for a natural value n if 1) it is true for n = 1 and 2) because it is valid for a natural value n = k, then it is true for n = k + 1 .

The method of mathematical induction is based on 3 stages:

1. We first check the accuracy of the output hardening at different natural values ​​of n (make the check work for one).
2. After which we check the validity for n = k.
3. And the validity of the statement has been demonstrated since n = k + 1.

How to implement the method of mathematical induction when dealing with inequalities and equalities

Let's take the butt we were talking about earlier.

Butt 1

Complete the formula S n = 1 1 · 2 + 1 2 · 3 +. . . + 1 n (n + 1) = n n + 1 .

Decision

As we already know, for the method of mathematical induction to work, it is necessary to complete three consecutive steps.

1. To begin with, we verify that the given equality will be fair for n equal to one. Derivable S 1 = 1 1 · 2 = 1 1 + 1 = 1 2 . Everything is correct here.
2. We further assume that the formula S k = k k + 1 is correct.
3. The third person needs to prove that S k + 1 = k + 1 k + 1 + 1 = k + 1 k + 2, based on the fairness of the former.

We can represent k + 1 as the sum of the first terms of the output sequence i k + 1:

S k + 1 = S k + 1 k + 1 (k + 2)

The fragments were removed in another step, so S k = k k + 1 can be written directly:

S k + 1 = S k + 1 k + 1 (k + 2).

Now the necessary changes have been completed. We need to visconate the reduced fraction to sleeping banner, bringing similar additions, formulate the formula for shortened multiplication and shorten those that came out:

S k + 1 = S k + 1 k + 1 (k + 2) = k k + 1 + 1 k + 1 (k + 2) = = k (k + 2) + 1 k + 1 (k + 2) = k 2 + 2 k + 1 k + 1 (k + 2) = (k + 1) 2 k + 1 (k + 2) = k + 1 k + 2

In this manner, we brought zeal to the third point, concluding all three steps to the method of mathematical induction.

Subject: Let's talk about the formula S n = n n + 1 є vernim.

The problems with trigonometric functions are much more complex.

Butt 2

Prove the identity of cos 2 α · cos 4 α · . . . · cos 2 n α = sin 2 n + 1 α 2 n sin 2 α.

Decision

As we remember, the first step is to check the accuracy of the equation at n, which is the same unit. To be clear, you need to know the basic trigonometric formulas.

cos 2 1 = cos 2 α sin 2 1 + 1 α 2 1 sin 2 α = sin 4 α 2 sin 2 α = 2 sin 2 α cos 2 α 2 sin 2 α = cos 2 α

Therefore, if n is equal to one, the identity will be true.

Now it is acceptable that justice can be preserved for n = k, then. It will be true that cos 2 α · cos 4 α · . . . · cos 2 k α = sin 2 k + 1 α 2 k sin 2 α.

The equality cos 2 α · cos 4 α · is brought to the fore. . . · cos 2 k + 1 α = sin 2 k + 2 α 2 k + 1 sin 2 α for the vipadku, if n = k + 1, taking the front subsumption as a basis.

Using the trigonometric formula,

sin 2 k + 1 α cos 2 k + 1 α = = 1 2 (sin (2 k + 1 α + 2 k + 1 α) + sin (2 k + 1 α - 2 k + 1 α)) = = 1 2 sin (2 2 k + 1 α) + sin 0 = 1 2 sin 2 k + 2 α

Otje,

cos 2 α · cos 4 α · . . . · cos 2 k + 1 α = = cos 2 α · cos 4 α · . . . · cos 2 k α · cos 2 k + 1 α = = sin 2 k + 1 α 2 k sin 2 α · cos 2 k + 1 α = 1 2 · sin 2 k + 1 α 2 k sin 2 α = sin 2 k + 2 α 2 k + 1 sin 2 α

An example of the greatest effort to prove the inequity of this method was found in the article about the method smallest squares. Read the paragraph where formulas are derived for finding approximation coefficients.

If you have marked a favor in the text, please see it and press Ctrl+Enter

In many branches of mathematics it is possible to prove the truth of the assertion, which lies in this way. truthfulness of love p(n) For " n ON (for whatever n ON p(n) true).

It is often necessary to bring using the method of mathematical induction.

It is based on the principle of mathematical induction. Therefore, it is chosen as one of the axioms of arithmetic and, therefore, is accepted without proof. Similar to the principle of mathematical induction, proposition p(n) The meaning of change is considered true for all natural substances, as two minds are drawn:

1. Proposition p(n) true for n= 1.

2. From the proposition that p(n) true for n =k (k- quite a natural number) trace, which is true for n =k+ 1.

Under the method of mathematical induction we understand this method of proof

1. Verify the truth of the confirmation for n= 1 – the basis of induction.

2. Assume that the firmness is correct for n = k - inductive allowance.

3. Argue that this is also correct for n =k+ 1 inductive junction.

Another proposition p(n) may not be true for all natural n, and starting to do something for n = n 0. And here the truth of the basis of induction is verified p(n) at n = n 0.

butt 1. Let it go. Bring it on

1. Induction base: at n= 1 for the previous ones S 1 = 1 and the formula has one result. Firmly verne.

n = k ta .

n = k+ 1. Let’s see what .

Effectively, through inductive assumption

This virus is resoluble

The inductive transition has been completed.

Respect. It’s good to write down what is given (inductive adjective) and what needs to be brought to light!

butt 2. Bring

1. The basis of induction. At n= 1, firmly, clearly, correctly.

2. Inductively not allowed. Let's go n = kі

3. Inductive transition. Let's go n = k+ 1. It is clear:

In fact, we know the right side of the square as the sum of two numbers:

The Vikorist inductive formula is assumed for the sum of arithmetic progression: , is rejected

butt 3. Bring up the nervousness

1. The basis of induction is the verification of the truth of the confirmation for, then. it is necessary to check the nervousness. For whom it is enough to know the inequality of the square: or 63< 64 – неравенство верно.

2. Let uneasiness be right for you.

3. Let it be clear:

Vikorist's assumption of induction

Knowing how the right side may look, this part of the imbalance is visible

It is difficult to establish that the exact multiplier does not outweigh the one. True,

butt 4. Show that for any natural number the number will end in a digit.

1. The least natural, for which a fair statement, is more ancient. .

2. Let the number end in . This means that this number can be written as if it were a natural number. Todi.

3. Let it go. Let's see that it ends with . Vikoristuyuchi otrimana vistavu, otrimaemo

The remaining number is equal to one.

supplement

1.4. Method of mathematical induction

As you can see, mathematical assertions (theorems) may be grounded and completed. We are familiar with one of the methods of proof - the method of mathematical induction.

In general, induction is a method of merging, which allows you to move from private to extraterrestrial ones. The gateway transition from the sacred to the private is called deduction.

Deduction will always lead to the correct conclusions. For example, we know the hidden result: all whole numbers that end in zero are divisible by 5. Of course, you can come up with a simple conclusion, such as a specific number that ends in 0, for example 18 0, divisible by 5.

At that very hour, induction can lead to incorrect conclusions. For example, noting that the number 60 is divisible by the numbers 1, 2, 3, 4, 5, 6, we have no right to write about those that 60 is divisible by any number.

The method of mathematical induction makes it possible, in many cases, to easily prove the validity of the formal statement P(n), the formulation of which includes the natural number n.

The method includes 3 stages.

1) Base of induction: we check the validity of the assertion P(n) for n = 1 (or another private value n, from which the validity of P(n) is transferred).

2) Assumption of induction: it is assumed that P(n) is valid for n = k.

3) Induction term: vikorist assumption, it is clear that P(n) is valid for n = k + 1.

As a result, one can come to an unambiguous conclusion about the validity of P(n) for any n ∈ N. Indeed, for n = 1 the statement is correct (base of induction). Also, it is true that n = 2, the fragments of transition from n = 1 to n = 2 primers (induction period). Due to the stagnation of the induction period of signs and signals, the validity of P(n) for n = 3, 4, 5, . . ., then P(n) is fair for all n.

Example 14. The sum of the first n unpaired natural numbers is equal to n2: 1 + 3 + 5 + …

+ (2n - 1) = n2.

Confirmation is carried out using the method of mathematical induction.

1) Base: with n=1 there is less than one addition, subtracting: 1 = 1.

Firmly verne.

2) Assumption: it is assumed that for any person k the following is true: 1 + 3 + 5 + … + (2k - 1) = k2.

Unraveling tasks about the reliability of falling under the hour of shooting

The zagalnaya setting of the plant is as follows:

The likelihood of exposure to a target per shot is $p$. $n$ shootings are carried out. Find the probability that the target will be hit exactly $k$ times (if $k$ hits).

The Bernoulli formula can be assumed and removed:

$$ P_n(k) = C_n^k \cdot p^k \cdot (1-p)^(n-k) = C_n^k \cdot p^k \cdot q^(n-k).

Here $C_n^k$ — the number goes from $n$ to $k$.

The problem involves a number of arrows with in different languages Hitting the target, the theory, the application of the solution and the calculator you can find here.

Video tutorial and Excel template

Watch our video about a new task about building in the Bernoulli scheme, find out how to use Excel to complete typical tasks.

You can freely download the Excel file from the video and use it to improve your tasks.

Apply the order to hit the target in a series of shots

Let's take a look at a number of typical stocks.

butt 1. There were 7 shootings. The rate of exposure per shot is equal to 0.705. Find out the certainty of the fact that with which there will be exactly 5 days.

It is clear that the problem involves repeated independent testing (shooting at a target), a total of $n=7$ shots are carried out, the probability of a hit with a cutaneous test is $p=0.705$, the probability of a miss is $q=1-p=1-0.70 5=$0.295 .

You need to calculate that there will be exactly $k=5$ hit. We put everything in formula (1) and remove it: $$ P_7(5)=C_(7)^5 \cdot 0.705^5 \cdot 0.295^2 = 21\cdot 0.705^5 \cdot 0.295^2= 0.318. $$

butt 2. The probability of exposure to the target per shot is 0.4.

Several independent shots were carried out on the target. Find out the certainty of what you would like to achieve for one purpose.

The most important parameters are: $n=4$ (after shooting), $p=0.4$ (event of exposure), $k \ge 1$ (if one would like to receive one).

Vikorist formula for the effectiveness of progesterone (no added water):

$$ P_4(k \ge 1) = 1-P_4(k \lt 1) = 1-P_4(0)= $$ $$ =1-C_(4)^0 \cdot 0.4^0 \cdot 0 .6^4 = 1-0.6^4 = 1-0.13 = 0.87. $$

I would like to spend one time out of almost 0.87 or 87%.

butt 3. The level of intensity of the archer's target remains at 0.3.

Know the likelihood that with six shots the target will be hit three to six times.

In advance of the task, it is necessary to know the probability of what number of events will occur in any given interval (and not exactly which number). Ale formula vikoristovuetsya excessively.

It is certain that the meta will be hit three to six times, so there will be either 3, or 4, or 5, or 6 hits.

These data are calculated using the formula (1):

$$P_6(3) = C_(6)^3\cdot 0.3^3\cdot 0.7^3 = 0.185. $$ $$ P_6(4)=C_(6)^4 \cdot 0.3^4\cdot 0.7^2 = 0.06. $$ $$ P_6(5)=C_(6)^5 \cdot 0.3^5\cdot 0.7^1 = 0.01. $$ $$ P_6(6)=C_(6)^6 \cdot 0.3^6\cdot 0.7^0 = 0.001.

The fragments of this nonsense, the incomprehensibility can be found using the formula for the addition of incomprehensibility: $$ P_6(3 \le k \le 6)=P_6(3)+P_6(4)+P_6(5)+P_6(6)=$$ $$ = 0.185+0.06+0.01+0.001=0.256.$$

butt 4. The reliability of one shot per target with four shots reaches 0.9984. Find out the probability of hitting the target in one shot.

The likelihood of hitting the target in one shot is significant. Let's introduce the following:
$A = $ (I would like to take one shot from the target),
and also the last word, which can be written as:
$\overline(A) = $ (All 4 shots will be for any purpose, as long as possible).

Let's write down the formula for the homovirality of $A$.

The given values ​​are: $ n = 4 $, $ P (A) = 0.9984 $. Substitute and remove from formula (1):

$$ P(A)=1-P(\overline(A))=1-P_4(0)=1-C_(4)^0 \cdot p^0 \cdot (1-p)^4=1- (1-p)^4 = 0.9984.

We believe that it is true that it happened:

$$1-(1-p)^4 = 0.9984, \(1-p)^4 = 0.0016, 1-p = 0.2, \p = 0.8. $$

Also, the probability of exposure to the target in one shot is equal to 0.8.

Whatever you read and share with others

Korisni poslannya

Find ready-made problems in the solution:

Online research on Bernoulli's formula

Viral inequality with the help of a calculator

Uneasiness in mathematics extends to all levels, where “=” is replaced by any of the following symbols: \ [> \] \ [\ geq \] \ [

* linear;

* Square;

* shotgun;

* indicative;

* trigonometric;

* logarithmic.

Therefore, due to this unevenness, they are called linear, partial, etc.

It is your responsibility to be aware of these signs:

* unevenness with a larger (>) or smaller symbol (

* Irregularities with icons such as greater or equal \[\geq\] lesser or equal [\leq\] are called unprofessional;

* the icon is not the same [[ne]] one, but it is necessary to consistently correct the problems with this icon.

Such instability exists in the appearance of a reversal of identities.

Also read our article “Beyond the solution for online dating”

It is acceptable that the uncertainty of the offensive is determined:

We believe that it is the same as linear alignment Be careful to follow the sign of unevenness.

From the beginning we transfer the members from the unknown to the left, from the visible to the right, changing symbols on the front:

Then we divide the offending parties by -4 and change the sign of inequality over time:

This is the testimony on the throne.

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The method of repeated mathematical induction

Decoupled ranks/Differential ranks

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Unraveling of differential levels

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Rivnyanya:

For additional calculator help, you can check differential comparison different foldability.

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Method of mathematical induction

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Main part

1. Full and uneven induction
2. Principle of mathematical induction
3. Method of mathematical induction
4. Butt solutions
5. Zealousness
6. Roster of numbers
7. Anxiety

Visnovok

List of Wikilists

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The basis of any mathematical investigation is deductive and inductive methods. The deductive method of merchandising is the merkuvannya from the hidden to the private, then. fading, the final moment of which is a hidden result, and the final moment is a private result. The induction stagnates during the transition from private results to secret ones, then. є by a method ranging from deductive.

The method of mathematical induction can be improved with progress. We start from the lowest, and as a result of logical thinking we arrive at the highest. The people have already given up progress, to develop their thoughts logically, since nature itself has designed it to grow inductively.

Although the scope of the method of mathematical induction has grown, the school curriculum devotes little time to it. Well, let’s say that the brown people should bring those two or three lessons, for which they will feel five words from the theory, learn five interesting tasks, and as a result, take away five for those that they know nothing.

It’s also so important - consider the dimensions inductively.

Main part

Behind its primary place, the word “induction” is stagnated until it disappears, after which it can be removed zagalini vysnovki, spiraling on the low private firmament. The simplest method of this kind of marting is repeated induction. The axle butt is so tarnished.

Please make sure that for a natural guy the number n is no more than 4< n < 20 представимо в виде суммы двух простых чисел. Для этого возьмём все такие числа и выпишем соответствующие разложения:

4=2+2; 6=3+3; 8=5+3; 10=7+3; 12=7+5;

14=7+7; 16=11+5; 18=13+5; 20=13+7.

These nine zeals show that the skin of the numbers that we need to dig up is effectively represented by the sum of two simple additions.

Also, there is a constant induction in the fact that the secret firmness is applied directly to the skin from the end number of possible episodes.

Sometimes the hidden result can be conveyed after looking at not everyone, but through a large number of roundabouts (this is what is called inconsistent induction).

The result, rejected by the induction, is lost, however, only by hypothesis, until the precise mathematical calculations are carried out, which surrounds everything around the fall. In other words, imperfect induction in mathematics is not respected by the legitimate method of strong proof, but rather by the strong method of discovering new truths.

For example, you need to know the sum of the first n consecutive unpaired numbers. Let's take a look around the fallout:

1+3+5+7+9=25=5 2

After looking at these many side-by-side outbursts, one is reminded of the onslaught of the following:

1+3+5+…+(2n-1)=n 2

tobto. the sum of the first n first unpaired numbers is equal to n 2

Of course, more caution can be a proof of the validity of the formula.

Repeated induction leads to less stagnation in mathematics. There are a lot of mathematical assertions to support an innumerable number of fallouts, and it is not possible for us to carry out verification for an uncountable number of fallouts. Improper induction often leads to mild results.

In many cases, the way out of such difficulties relies on a special method of induction, called the method of mathematical induction. Vіn lies at the next step.

It is necessary to prove the validity of this assertion for any natural number n (for example, it is necessary to prove that the sum of the first n unpaired numbers is equal to n 2). A complete re-verification of this firmness of the skin value n is impossible, since the number of natural numbers is endless. To complete the conclusion, first check its validity for n=1. Then we argue that any natural value k from the justice of the considered statement for n=k receives its justice for n=k+1.

This affirmation is respected and communicated to everyone. True, the statement is fair for n=1. Ale todi vono faire y odnogo number n=1+1=2. This fairness of the affirmation for n=2 results in the same fairness for n=2+

1=3. This shows the validity of the confirmation for n=4, etc. It dawned on us that, finally, we will come to any natural number n. Well, the affirmation is more correct for any n.

Having already said this, let us formulate the offensive principle behind the fire.

The principle of mathematical induction.

Since the proposition A(n), which lies with the natural number n, is true for n=1 and because it is true for n=k (if k is a natural number), then it is true for the natural number n=k +1, the assumption A(n) is true for any natural number n.

In a number of cases, it is necessary to prove the validity of this assertion not for all natural numbers, but especially for n>p, where p is a fixed natural number. And here the principle of mathematical induction is formulated as follows.

If the proposition A(n) is true for n=p and if A(k)ÞA(k+1) for any k>p, then the proposition A(n) is true for any n>p.

Proof by the method of mathematical induction is carried out in this way. From now on the firmness must be verified for n=1, then. the truth of condition A(1) is established. This part of the confirmation is called the basis of induction. Then comes the part of the proof called the induction trial. Whose part is to prove the validity of the assertion for n=k+1 and the assumed validity of the assertion for n=k (assumed induction), then. prove that A(k) ÞA(k+1).

Bring that 1+3+5+…+(2n-1)=n 2.

Solution: 1) May n=1=1 2 . Otje,

the assertion is correct for n=1, then. A(1) is true.

2) Let us prove that A(k) ÞA(k+1).

Let k be a natural number and let the affirmation be true for n = k, then.

1+3+5+…+(2k-1)=k 2 .

Let us prove that the assertion is also valid for the advancing natural number n=k+1, then. what

1+3+5+…+(2k+1)=(k+1) 2 .

True,

1+3+5+…+(2k-1)+(2k+1)=k 2 +2k+1=(k+1) 2 .

Otzhe, A(k) ÞA(k+1). According to the principle of mathematical induction, the assumption A(n) is truly nÎN.

Bring it on

1+x+x 2 +x 3 +…+x n =(x n+1 -1)/(x-1), de x1

Solution: 1) When n=1 it is eliminated

1+x=(x 2 -1)/(x-1)=(x-1)(x+1)/(x-1)=x+1

Also, for n=1 the formula is true; A(1) is true.

2) Let k be a natural number and let the formula be correct for n = k, then.

1+x+x 2 +x 3 +…+x k =(x k+1 -1)/(x-1).

Let us see that this is where jealousy comes to an end.

1+x+x 2 +x 3 +…+x k +x k+1 =(x k+2 -1)/(x-1).

True

1+x+x 2 +x 3 +…+x k +x k+1 =(1+x+x 2 +x 3 +…+x k)+x k+1 =

=(x k+1 -1)/(x-1)+x k+1 =(x k+2 -1)/(x-1).

Otzhe, A(k) ÞA(k+1). Based on the principle of mathematical induction, it follows that the formula is correct for any natural number n.

Bring that the number of diagonals of a convex n-kutnik is equal to n(n-3)/2.

Solution: 1) At n=3 the right

And 3 wisely, more in trikutnik

A 3 =3(3-3)/2=0 diagonals;

A 2 A(3) is true.

2) It is acceptable that everyone

the bulging k-kutnik bears

A 1 x A k = k (k-3) / 2 diagonals.

And k Let’s see what’s wrong with the bulging

(k+1)-kutnik number

diagonals A k+1 =(k+1)(k-2)/2.

Nehai A 1 A 2 A 3 …A k A k +1 - opukly (k + 1)-kutnik. Let's draw the new diagonal A1Ak. To adjust the number of diagonals of this (k+1)-gon, you need to adjust the number of diagonals in the k-box A 1 A 2 ...A k , add to the calculated number k-2, then. the number of diagonals of the (k+1)-kutnik that extend from the vertices A k+1 i, in addition, follow the diagonal A 1 A k .

In such a manner

 k+1 =  k +(k-2)+1=k(k-3)/2+k-1=(k+1)(k-2)/2.

Otzhe, A(k) ÞA(k+1). Following the principle of mathematical induction, the assertion is correct for any convex n-column.

To convey that for anyone there is a fair affirmation:

1 2 +2 2 +3 2 +…+n 2 =n(n+1)(2n+1)/6.

Resolution: 1) Let n = 1, then

X 1 = 1 2 = 1 (1 +1) (2 +1) / 6 = 1.

Well, with n=1 the firmament is more correct.

2) It is acceptable that n=k

X k =k 2 =k(k+1)(2k+1)/6.

3) Let’s look at the given solidification for n=k+1

X k+1 = (k+1)(k+2)(2k+3)/6.

X k+1 =1 2 +2 2 +3 2 +…+k 2 +(k+1) 2 =k(k+1)(2k+1)/6+ +(k+1) 2 =(k (k+1)(2k+1)+6(k+1) 2)/6=(k+1)(k(2k+1)+

6(k+1))/6=(k+1)(2k 2 +7k+6)/6=(k+1)(2(k+3/2)(k+

2))/6=(k+1)(k+2)(2k+3)/6.

We have established the validity of equality i for n = k + 1, and, due to the method of mathematical induction, the statement is more correct for any natural number n.

To convey that for any natural n jealousy is fair:

1 3 +2 3 +3 3 + ... + n 3 = n 2 (n + 1) 2 /4.

Resolution: 1) Let n = 1.

Todi X 1 = 13 = 12 (1 +1) 2 / 4 = 1.

Mi, what is n=1 is more correct.

2) It is acceptable that jealousy is correct for n = k

X k = k 2 (k +1) 2/4.

3) Let us prove the truth of this assertion for n=k+1, then.

X k+1 =(k+1) 2 (k+2) 2/4. X k+1 =1 3 +2 3 +…+k 3 +(k+1) 3 =k 2 (k+1) 2 /4+(k+1) 3 =(k 2 (k++1) 2 +4(k+1) 3)/4=(k+1) 2 (k 2 +4k+4)/4=(k+1) 2 (k+2) 2 /4.

From the above proof it is clear that the assertion is true for n = k + 1, therefore, jealousy is true for any natural number n.

Bring it on

((2 3 +1)/(2 3 -1))´((3 3 +1)/(3 3 -1))´…´((n 3 +1)/(n 3 -1))= 3n(n+1)/2(n 2 +n+1), where n>2.

Solution: 1) For n=2 the identity looks like: (2 3 +1)/(2 3 -1)=(3´2´3)/2(2 2 +2+1),

tobto. That's right.

2) It is acceptable that the expression is correct for n = k

(2 3 +1)/(2 3 -1)´…´(k 3 +1)/(k 3 -1)=3k(k+1)/2(k 2 +k+1).

3) Let us prove the correctness of the virus at n=k+1.

(((2 3 +1)/(2 3 -1))´…´((k 3 +1)/(k 3 -1))))(((k+1) 3 +

1)/((k+1) 3 -1))=(3k(k+1)/2(k 2 +k+1))´((k+2)((k+

1) 2 -(k+1)+1)/k((k+1) 2 +(k+1)+1))=3(k+1)(k+2)/2´

´((k+1) 2 +(k+1)+1).

We have established the validity of equality for n=k+1, then, through the method of mathematical induction, the statement is correct for any n>2

Bring it on

1 3 -2 3 +3 3 -4 3 +…+(2n-1) 3 -(2n) 3 =-n 2 (4n+3)

for any natural n.

Resolution: 1) Let n = 1, then

1 3 -2 3 =-1 3 (4+3); -7=-7.

2) It is acceptable that n=k then

1 3 -2 3 +3 3 -4 3 +…+(2k-1) 3 -(2k) 3 =-k 2 (4k+3).

3) Let us prove the truth of this assertion at n=k+1

(1 3 -2 3 +…+(2k-1) 3 -(2k) 3)+(2k+1) 3 -(2k+2) 3 =-k 2 (4k+3)+

+(2k+1) 3 -(2k+2) 3 =-(k+1) 3 (4(k+1)+3).

It has been proven that the equation is true for n=k+1, so the assertion is more correct for any natural n.

Make sure the sameness is true

(1 2 /1´3)+(2 2 /3´5)+…+(n 2 /(2n-1)´(2n+1))=n(n+1)/2(2n+1)

for any natural n.

1) For n = 1, the identity is true: 1 2 /1 '3 = 1 (1 +1) / 2 (2 +1).

2) It is acceptable that n=k

(1 2 /1´3)+…+(k 2 /(2k-1)´(2k+1))=k(k+1)/2(2k+1).

3) Let us prove that the identity is correct for n=k+1.

(1 2 /1´3)+…+(k 2 /(2k-1)(2k+1))+(k+1) 2 /(2k+1)(2k+3)=(k(k+ 1 )/2(2k+1))+((k+1) 2 /(2k+1)(2k+3))=((k+1)/(2k+1))´((k/2 ) +((k+1)/(2k+3)))=(k+1)(k+2)´ (2k+1)/2(2k+1)(2k+3)=(k+1 ) (k+2)/2(2(k+1)+1).

From the above evidence it is clear that the firmness is true for any natural n.

Bring that (11n+2+122n+1) is divisible by 133 without excess.

Resolution: 1) Let n = 1, then

11 3 +12 3 = (11 +12) (11 2 -132 +12 2) = 23 '133.

Ale (23´133) is divided by 133 without excess, so with n=1 the firmness is true; A(1) is true.

2) It is acceptable that (11 k+2 +12 2k+1) is divided by 133 without excess.

3) Let us explain what this time has

(11 k+3 +12 2k+3) is divided by 133 without excess. True 11 k+3 +12 2k+3 =11´11 k+2 +12 2´ 12 2k+1 =11´11 k+2 +

+(11+133)´12 2k+1 =11(11 k+2 +12 2k+1)+133´12 2k+1 .

The sum is rejected to be divided by 133 without excess, since the first addition is divided by 133 without excess for allowances, and in the other one of the multipliers is 133. So, A(k) ÞA(k+1). The strength of the method of mathematical induction has been achieved.

Bring that for any n 7 n -1 is divisible by 6 without excess.

Solution: 1) Let n = 1, then X 1 = 7 1 -1 = 6 be divided by 6 without excess. This means that for n = 1 the statement is true.

2) It is acceptable that n=k

7 k-1 is divided by 6 without excess.

3) Let us prove that the assertion is valid for n=k+1.

X k +1 = 7 k +1 -1 = 7 '7 k -7 +6 = 7 (7 k -1) +6.

The first addition is divided by 6, the fragments 7 k -1 are divided by 6 for the soup, and the other addition is 6. So 7 n -1 is a multiple of 6 for any natural number n. The strength of the method of mathematical induction has been achieved.

Bring that 3 3n-1 +2 4n-3 with sufficient natural n is divisible by 11.
Resolution: 1) Let n = 1, then

X 1 = 3 3-1 +2 4-3 = 3 2 +2 1 = 11 divided by 11 without excess. Well, with n=1 the firmament is more correct.

2) It is acceptable that n=k

X k =3 3k-1+24k-3 is divided by 11 without excess.

3) Let us prove that the assertion is correct for n=k+1.

X k+1 =3 3(k+1)-1 +2 4(k+1)-3 =3 3k+2 +2 4k+1 =3 3´ 3 3k-1 +2 4´ 2 4k-3 =

27´3 3k-1 +16´2 4k-3 =(16+11)´3 3k-1 +16´2 4k-3 =16´3 3k-1 +

11´3 3k-1 +16´2 4k-3 =16(3 3k-1 +2 4k-3)+11´3 3k-1 .

First, the additional sum is divided by 11 without excess, the fragments 3 3k-1 +2 4k-3 are divided by 11 after allowances, the other is divided by 11, since one of its multipliers is the number 11. This means that the sum is divided by 11 without excess for be- any natural n. The strength of the method of mathematical induction has been achieved.

Bring that 11 2n -1 with sufficient natural n is divisible by 6 without excess.

Resolution: 1) Let n=1, then 112-1=120 be divided by 6 without excess. This means that for n = 1 the statement is true.

2) It is acceptable that n=k

11 2k -1 is divided by 6 without excess.

11 2(k+1) -1=121´11 2k -1=120´11 2k +(11 2k -1).

The offense of additions is divided by 6 without excess: first, the number is divisible by 6, the number is 120, and the other is divided by 6 without excess for soups. This means that the sum is divisible by 6 without excess. The method of mathematical induction has been proven.

Bring that 3 3n+3 -26n-27 with sufficient natural n is divided by 26 2 (676) without excess.

Solution: Let us prove in advance that 33n+3-1 is divisible by 26 without excess.

1. When n=0

3 3 -1=26 is divided by 26

2. It is acceptable that n=k

3 3k+3 -1 is divided by 26

3. Let us know what is firmly established

Correct for n=k+1.

3 3k+6 -1=27´3 3k+3 -1=26´3 3l+3 +(3 3k+3 -1) – extend by 26

Now we will prove the affirmation formulated in the mind of the master.

1) Obviously, for n = 1 the statement is true

3 3+3 -26-27=676

2) It is acceptable that n=k

Viraz 3 3k+3 -26k-27 is divided by 26 2 without excess.

3) Let us prove that the assertion is correct for n=k+1

3 3k+6 -26(k+1)-27=26(3 3k+3 -1)+(3 3k+3 -26k-27).

The resentment of the dodanks is divided by 26 2; First, divide by 26 2, so that we have brought the completeness to 26 degrees, which stands at the arms, and the other is divided by induction. Based on the method of mathematical induction, the conclusion has been reached.

Prove that if n>2 and x>0, then inequality is true

(1+x) n >1+n´x.

Solution: 1) When n=2 inequality is fair, only

(1+x) 2 = 1+2x+x 2 >1+2x.

Well, A(2) is true.

2) Let us prove that A(k) ÞA(k+1), since k> 2. It is admissible that A(k) is true, then inequality is true

(1+x) k >1+k´x. (3)

Let us prove that even A(k+1) is true, so that inequality is true

(1+x) k+1 >1+(k+1)´x.

In fact, multiplying the offending parts of inequality (3) by the positive number 1+x is eliminated

(1+x) k+1 >(1+k´x)(1+x).

Let's look at the right part of the remaining nervous-

stva; maєmo

(1+k´x)(1+x)=1+(k+1)´x+k´x 2 >1+(k+1)´x.

The result is clear that

(1+x) k+1 >1+(k+1)´x.

Otzhe, A(k) ÞA(k+1). On the basis of the principle of mathematical induction, it can be confirmed that Bernoulli’s inequality is valid for any

Convey that the uneasiness is fair

(1+a+a 2) m > 1+m'a+(m(m+1)/2)'a 2 for a> 0.

Solution: 1) When m=1

(1+a+a 2) 1 > 1+a+(2/2)´a 2 offending parts of equal.

2) It is acceptable that m=k

(1+a+a 2) k >1+k'a+(k(k+1)/2)'a 2

3) Let us prove that with m=k+1 inequality is correct

(1+a+a 2) k+1 =(1+a+a 2)(1+a+a 2) k >(1+a+a 2)(1+k+a+)

+(k(k+1)/2)´a 2)=1+(k+1)´a+((k(k+1)/2)+k+1)´a 2 +

+((k(k+1)/2)+k)´a 3 +(k(k+1)/2)´a 4 > 1+(k+1)´a+

+((k+1)(k+2)/2)'a 2 .

We have established the validity of inequality for m=k+1, and then, by virtue of the method of mathematical induction, inequality is valid for any natural m.

Show that for n>6 inequality is true

3 n >n´2 n+1 .

Resolution: Rewrite the nervousness in your appearance

1. At n=7 it is possible

3 7 /2 7 = 2187/128> 14 = 2 '7

The anxiety is true.

2. It is acceptable that n=k

3) Let us show the accuracy of the inequality at n = k +1.

3 k+1 /2 k+1 =(3 k /2 k)'(3/2)>2k'(3/2)=3k>2(k+1).

Fragments of k>7, remaining nervousness is obvious.

By virtue of the method of mathematical induction, inequality is true for any natural number n.

Show that for n>2 inequality is true

1+(1/2 2)+(1/3 2)+…+(1/n 2)<1,7-(1/n).

Solution: 1) When n=3 the inequality is true

1+(1/2 2)+(1/3 2)=245/180<246/180=1,7-(1/3).

1. It is acceptable that n=k

1+(1/2 2)+(1/3 2)+…+(1/k 2)=1.7-(1/k).

3) Let’s bring justice to non-

equal for n=k+1

(1+(1/2 2)+…+(1/k 2))+(1/(k+1) 2)<1,7-(1/k)+(1/(k+1) 2).

Let's see that 1.7-(1/k)+(1/(k+1) 2)<1,7-(1/k+1)Û

û(1/(k+1) 2)+(1/k+1)<1/kÛ(k+2)/(k+1) 2 <1/kÛ

K(k+2)<(k+1) 2Û k 2 +2k

It remains obvious, but

1+(1/2 2)+(1/3 2)+…+(1/(k+1) 2)<1,7-(1/k+1).

Looking at the method of mathematical induction, inequality has been achieved.

Visnovok

Having learned the method of mathematical induction, I advanced my knowledge in this field of mathematics, and also learned to overcome problems that were previously beyond my power.

This has logical and purposeful tasks, then. the very ones that promote interest to mathematics itself, as well as to science. The versatility of such tasks becomes useful employment and you can get more and more interesting ones in mathematical labs. In my opinion, this is the basis of any science.

By continuing to learn the method of mathematical induction, I will learn to understand this not only in mathematics, but also in the most important problems of physics, chemistry and life itself.

MATHEMATICS:

LECTURES, TEACHING, DECISION

Basic publication / V. G. Boltyansky, Yu. V. Sidorov, M. I. Shabunin. TOV "Popuri" 1996.

ALGEBRA AND A BACK TO ANALYSIS

Basic handbook / I. T. Demidov, A. N. Kolmogorov, S. I. Schwarzburg, O. S. Ivashev-Musatov, B. E. Weitz. "Osvita" 1975.