Method of mathematical induction. Its application to solving algebraic problems Solving differential equations

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Introduction

This topic is relevant, since every day people solve various problems in which they use different solution methods, but there are tasks in which one cannot do without the method of mathematical induction, and in such cases knowledge in this area will be very useful.

I chose this topic for research because little time is devoted to the method of mathematical induction in the school curriculum; the student learns superficial information that will only help him obtain general idea about this method, but to study this theory in depth, self-development will be required. It will really be useful to learn more about this topic, as it broadens a person’s horizons and helps in solving complex problems.

Goal of the work:

Get acquainted with the method of mathematical induction, systematize knowledge on this topic and apply it when solving mathematical problems and proving theorems, justify and clearly show the practical significance of the method of mathematical induction as a necessary factor for solving problems.

Job objectives:

Analyze the literature and summarize knowledge on this topic.

Understand the principle of the method of mathematical induction.

Explore the application of the method of mathematical induction to problem solving.

Formulate conclusions and conclusions on the work done.

Main part of the study

History:

Only towards the end of the 19th century did a standard of requirements for logical rigor emerge, which remains dominant in modern literature to this day. practical work mathematicians on the development of individual mathematical theories.

Induction is a cognitive procedure through which a statement generalizing them is derived from a comparison of existing facts.

In mathematics, the role of induction is largely that it underlies the chosen axiomatics. After long-term practice showed that a straight path is always shorter than a curved or broken one, it was natural to formulate an axiom: for any three points A, B and C, an inequality holds.

The awareness of the method of mathematical induction as a separate important method goes back to Blaise Pascal and Gersonides, although individual cases of application are found in ancient times in Proclus and Euclid. Modern name The method was introduced by De Morgan in 1838.

The method of mathematical induction can be compared to progress: we start from the lowest, and as a result of logical thinking we come to the highest. Man has always strived for progress, for the ability to logically develop his thoughts, which means that nature itself destined him to think inductively.

Induction and deduction

It is known that there are both particular and general statements, and these two terms are based on the transition from one to the other.

Deduction (from Latin deductio - deduction) - a transition in the process of cognition from general knowledge to private And single. In deduction, general knowledge serves as the starting point of reasoning, and this general knowledge is assumed to be “ready-made,” existing. The peculiarity of deduction is that the truth of its premises guarantees the truth of the conclusion. Therefore, deduction has enormous persuasive power and is widely used not only to prove theorems in mathematics, but also wherever reliable knowledge is needed.

Induction (from Latin inductio - guidance) is a transition in the process of cognition from private knowledge to general In other words, it is a method of research and cognition associated with generalizing the results of observations and experiments. A feature of induction is its probabilistic nature, i.e. If the initial premises are true, the conclusion of induction is only probably true and in the final result it can turn out to be either true or false.

Complete and incomplete induction

Inductive inference is a form of abstract thinking in which thought develops from knowledge to a lesser extent generality to knowledge of a greater degree of generality, and the conclusion arising from the premises is predominantly probabilistic in nature.

During the research, I found out that induction is divided into two types: complete and incomplete.

Complete induction is an inference in which a general conclusion about a class of objects is made based on the study of all objects of this class.

For example, let it be necessary to establish that every even natural number n within the range 6≤ n≤ 18 can be represented as the sum of two prime numbers. To do this, take all such numbers and write out the corresponding expansions:

6=3+3; 8=5+3; 10=7+3; 12=7+5;14=7+7; 16=11+5; 18=13+5;

These equalities show that each of the numbers we are interested in is indeed represented as the sum of two simple terms.

Consider the following example: sequence yn= n 2 +n+17; Let's write out the first four terms: y 1 =19; y 2 =23; y 3 =29; y 4 =37; Then we can assume that the entire sequence consists of prime numbers. But this is not so, let's take y 16 = 16 2 +16+17=16(16+1)+17=17*17. This is a composite number, which means our assumption is incorrect, thus, incomplete induction does not lead to completely reliable conclusions, but allows us to formulate a hypothesis, which subsequently requires mathematical proof or refutation.

Method of mathematical induction

Complete induction has only limited applications in mathematics. Many interesting mathematical statements cover an infinite number of special cases, and we are not able to test for all these situations. But how can we test for an infinite number of cases? This method was proposed by B. Pascal and J. Bernoulli, this is a method of mathematical induction, which is based on principle of mathematical induction.

If a sentence A(n), depending on a natural number n, is true for n=1 and from the fact that it is true for n=k (where k is any natural number), it follows that it is also true for the next number n=k +1, then assumption A(n) is true for any natural number n.

In a number of cases, it may be necessary to prove the validity of a certain statement not for all natural numbers, but only for n>p, where p is a fixed natural number. In this case, the principle of mathematical induction is formulated as follows:

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

Algorithm (it consists of four stages):

1.base(we show that the statement being proved is true for some simplest special cases ( P = 1));

2.assumption(we assume that the statement has been proven for the first To cases); 3 .step(under this assumption we prove the statement for the case P = To + 1); 4.output (at the statement is true for all cases, that is, for all P) .

Note that the method of mathematical induction can not solve all problems, but only problems parameterized by a certain variable. This variable is called the induction variable.

Application of the method of mathematical induction

Let's apply this entire theory in practice and find out in what problems this method is used.

Problems to prove inequalities.

Example 1. Prove Bernoulli's inequality(1+x)n≥1+n x, x>-1, n € N.

1) For n=1 the inequality is true, since 1+x≥1+x

2) Suppose that the inequality is true for some n=k, i.e.

(1+x) k ≥1+k x.

Multiplying both sides of the inequality by a positive number 1+x, we get

(1+x) k+1 ≥(1+kx)(1+ x) =1+(k+1) x + kx 2

Taking into account that kx 2 ≥0, we arrive at the inequality

(1+x) k+1 ≥1+(k+1) x.

Thus, from the assumption that Bernoulli's inequality is true for n=k, it follows that it is true for n=k+1. Based on the method of mathematical induction, it can be argued that Bernoulli’s inequality is valid for any n € N.

Example 2. Prove that for any natural number n>1, .

Let's prove it using the method of mathematical induction.

Let us denote the left side of the inequality by.

1), therefore, for n=2 the inequality is valid.

2) Let for some k. Let us prove that then and. We have, .

Comparing and, we have, i.e. .

For any positive integer k, the right-hand side of the last equality is positive. That's why. But that means and. We have proven the validity of the inequality for n=k+1, therefore, by virtue of the method of mathematical induction, the inequality is valid for any natural number n>1.

Problems to prove identities.

Example 1. Prove that for any natural number n the equality is true:

1 3 +2 3 +3 3 +…+n 3 =n 2 (n+1) 2 /4.

Let n=1, then X 1 =1 3 =1 2 (1+1) 2 /4=1.

We see that for n=1 the statement is true.

2) Suppose that the equality is true for n=kX k =k 2 (k+1) 2 /4.

3) Let us prove the truth of this statement for n=k+1, i.e. 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 statement is true for n=k+1, therefore, the equality is true for any natural number n.

Example 2. Prove that for any natural n the equality is true

1) Let us check that this identity is true for n = 1.; - right.

2) Let the identity also be true for n = k, i.e..

3) Let us prove that this identity is also true for n = k + 1, i.e.;

Because If the equality is true for n=k and n=k+1, then it is true for any natural number n.

Summation problems.

Example 1. Prove that 1+3+5+…+(2n-1)=n 2.

Solution: 1) We have n=1=1 2 . Therefore, the statement is true for n=1, i.e. A(1) is true.

2) Let us prove that A(k) A(k+1).

Let k be any natural number and let the statement be true for n=k, i.e. 1+3+5+…+(2k-1)=k 2 .

Let us prove that then the statement is also true for the next natural number n=k+1, i.e. What

1+3+5+…+(2k+1)=(k+1) 2 .

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

So, A(k) A(k+1). Based on the principle of mathematical induction, we conclude that assumption A(n) is true for any n N.

Example 2. Prove the formula, n is a natural number.

Solution: When n=1, both sides of the equality turn to one and, therefore, the first condition of the principle of mathematical induction is satisfied.

Let's assume that the formula is correct for n=k, i.e. .

Let's add to both sides of this equality and transform the right side. Then we get

Thus, from the fact that the formula is true for n=k, it follows that it is also true for n=k+1, then this statement is true for any natural number n.

Divisibility problems.

Example 1. Prove that (11 n+2 +12 2n+1) is divisible by 133 without a remainder.

Solution: 1) Let n=1, then

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

(23×133) is divisible by 133 without a remainder, which means that for n=1 the statement is true;

2) Suppose that (11 k+2 +12 2k+1) is divisible by 133 without a remainder.

3) Let us prove that in this case

(11 k+3 +12 2k+3) is divisible by 133 without a remainder. Indeed, 11 k+3 +12 2l+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 resulting sum is divided by 133 without a remainder, since its first term is divisible by 133 without a remainder by assumption, and in the second one of the factors is 133.

So, A(k)→ A(k+1), then based on the method of mathematical induction, the statement is true for any natural n.

Example 2. Prove that 3 3n-1 +2 4n-3 for an arbitrary natural number n is divisible by 11.

Solution: 1) Let n=1, then X 1 =3 3-1 +2 4-3 =3 2 +2 1 =11 is divisible by 11 without a remainder. This means that for n=1 the statement is true.

2) Suppose that for n=k

X k =3 3k-1 +2 4k-3 is divisible by 11 without a remainder.

3) Let us prove that the statement is true 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 .

The first term is divisible by 11 without a remainder, since 3 3k-1 +2 4k-3 is divisible by 11 by assumption, the second is divisible by 11, because one of its factors is the number 11. This means that the sum is divisible by 11 without a remainder for any natural number n.

Problems from real life.

Example 1. Prove that the sum Sn of interior angles of any convex polygon is equal to ( P- 2)π, where P— the number of sides of this polygon: Sn = ( P- 2)π (1).

This statement does not make sense for all natural P, but only for P > 3, since the minimum number of angles in a triangle is 3.

1) When P= 3 our statement takes the form: S 3 = π. But the sum of the interior angles of any triangle is indeed π. Therefore, when P= 3 formula (1) is correct.

2) Let this formula be true for n =k, that is S k = (k- 2)π, where k > 3. Let us prove that in this case the formula holds: S k+ 1 = (k- 1)π.

Let A 1 A 2 ... A k A k+ 1—arbitrary convex ( k+ 1) -gon (Fig. 338).

Connecting points A 1 and A k , we get convex k-gon A 1 A 2 ... A k — 1 A k . Obviously, the sum of the angles ( k+ 1) -gon A 1 A 2 ... A k A k+ 1 is equal to the sum of the angles k-gon A 1 A 2 ... A k plus the sum of the angles of a triangle A 1 A k A k+ 1 . But the sum of the angles k-gon A 1 A 2 ... A k by assumption equal to ( k- 2)π, and the sum of the angles of the triangle A 1 A k A k+ 1 is equal to π. That's why

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

So, both conditions of the principle of mathematical induction are satisfied, and therefore formula (1) is true for any natural P > 3.

Example 2. There is a staircase, all steps of which are the same. It is required to indicate the minimum number of positions that would guarantee the ability to “climb” onto any step by number.

Everyone agrees that there must be a condition. We must be able to climb to the first step. Next, they must be able to climb from the 1st step to the second. Then to the second - to the third, etc. to the nth step. Of course, in totality, “n” statements guarantee that we will be able to get to the nth step.

Let's now look at the 2, 3,..., n position and compare them with each other. It is easy to see that they all have the same structure: if we have reached the k step, then we can climb up to the (k+1) step. Hence, the following axiom becomes natural for the validity of statements depending on “n”: if a sentence A(n), in which n is a natural number, holds for n=1 and from the fact that it holds for n=k (where k is any natural number), it follows that it holds for n=k+1, then assumption A(n) holds for any natural number n.

Application

Problems using the method of mathematical induction when entering universities.

Note that upon admission to higher education educational establishments There are also problems that can be solved by this method. Let's look at them using specific examples.

Example 1. Prove that any natural P equality is true

1) When n=1 we get the correct equality Sin.

2) Having made the induction assumption that when n= k the equality is true, consider the sum on the left side of the equality for n =k+1;

3) Using reduction formulas, we transform the expression:

Then, by virtue of the method of mathematical induction, the equality is true for any natural number n.

Example 2. Prove that for any natural number n the value of the expression 4n +15n-1 is a multiple of 9.

1) With n=1: 2 2 +15-1=18 - a multiple of 9 (since 18:9=2)

2) Let the equality hold for n=k: 4 k +15k-1 multiple of 9.

3) Let us prove that the equality holds for the next number n=k+1

4 k+1 +15(k+1)-1=4 k+1 +15k+15-1=4.4 k +60k-4-45k+18=4(4 k +15k-1)-9(5k- 2)

4(4 k +15k-1) - multiple of 9;

9(5k-2) - multiple of 9;

Consequently, the entire expression 4(4 k +15k-1)-9(5k-2) is a multiple of 9, which is what needed to be proven.

Example 3. Prove that for any natural number P the condition is met: 1∙2∙3+2∙3∙4+…+ p(p+1)(p+2)=.

1) Let's check that this formula is correct when n=1: Left side = 1∙2∙3=6.

Right part = . 6 = 6; true when n=1.

2) Suppose that this formula is true for n =k:

1∙2∙3+2∙3∙4+…+k(k+1)(k+2)=. S k =.

3) Let us prove that this formula is true for n =k+1:

1∙2∙3+2∙3∙4+…+(k+1)(k+2)(k+3)=.

S k+1 =.

Proof:

So, this condition is true in two cases and has been proven to be true for n =k+1, therefore it is true for any natural number P.

Conclusion

To summarize, in the process of research I found out what induction is, which can be complete or incomplete, got acquainted with the method of mathematical induction based on the principle of mathematical induction, and considered many problems using this method.

I also learned a lot of new information, different from what is included in the school curriculum. While studying the method of mathematical induction, I used various literature, Internet resources, and also consulted with a teacher.

Conclusion: Having generalized and systematized knowledge on mathematical induction, I became convinced of the need for knowledge on this topic in reality. Positive quality The method of mathematical induction is its wide application in solving problems: in the field of algebra, geometry and real mathematics. This knowledge also increases interest in mathematics as a science.

I am confident that the skills I acquired during my work will help me in the future.

​ Bibliography

Sominsky I.S. Method of mathematical induction. Popular lectures on mathematics, issue 3-M.: Science, 1974.

L. I. Golovina, I. M. Yaglom. Induction in geometry. - Fizmatgiz, 1961. - T. 21. - 100 p. — (Popular lectures on mathematics).

Dorofeev G.V., Potapov M.K., Rozov N.Kh. A manual on mathematics for those entering universities (Selected questions of elementary mathematics) - 5th edition, revised, 1976 - 638 pp.

A. Shen. Mathematical induction. - MCNMO, 2004. - 36 p.

M.L. Galitsky, A.M. Goldman, L.I. Zvavich Collection of problems in algebra: textbook for grades 8-9. with depth studying mathematics 7th ed. - M.: Prosveshchenie, 2001. - 271 p.

Ma-ka-ry-chev Yu.N., Min-dyuk N.G Additional chapters for the school textbook of al-geb-ry 9th grade. - M.: Pro-sve-shche-nie, 2002.

Wikipedia is a free encyclopedia.

Introduction

Main part

1. Complete and incomplete induction

2. Principle of mathematical induction

3. Method of mathematical induction

4. Solving examples

5. Equalities

6. Dividing numbers

7. Inequalities

Conclusion

List of used literature

Introduction

The basis of any mathematical research is deductive and inductive methods. The deductive method of reasoning is reasoning from the general to the specific, i.e. reasoning, the starting point of which is the general result, and the final point is the particular result. Induction is used when moving from particular results to general ones, i.e. is the opposite of deductive method.

The method of mathematical induction can be compared to progress. We start from the lowest, and as a result of logical thinking we come to the highest. Man has always strived for progress, for the ability to develop his thoughts logically, which means that nature itself destined him to think inductively.

Although the scope of application of the method of mathematical induction has grown, little time is devoted to it in the school curriculum. Well, say what useful to a person will bring those two or three lessons, during which he will hear five words of theory, solve five primitive problems, and, as a result, will receive an A for the fact that he knows nothing.

But it is so important to be able to think inductively.

Main part

In its original meaning, the word “induction” is applied to reasoning through which general conclusions are obtained based on a number of specific statements. The simplest method of reasoning of this kind is complete induction. Here is an example of such reasoning.

Let it be necessary to establish that every even natural number n within 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 equalities show that each of the numbers we are interested in is indeed represented as the sum of two simple terms.

Thus, complete induction consists of proving the general statement separately in each of a finite number of possible cases.

Sometimes the general result can be predicted after considering not all, but a sufficiently large number of particular cases (the so-called incomplete induction).

The result obtained by incomplete induction remains, however, only a hypothesis until it is proven by precise mathematical reasoning, covering all special cases. In other words, incomplete induction in mathematics is not considered a legitimate method of rigorous proof, but is a powerful method for discovering new truths.

Let, for example, you want to find the sum of the first n consecutive odd numbers. Let's consider special cases:

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

After considering these few special cases, the following general conclusion suggests itself:

1+3+5+…+(2n-1)=n 2

those. the sum of the first n consecutive odd numbers is n 2

Of course, the observation made cannot yet serve as proof of the validity of the given formula.

Complete induction has only limited applications in mathematics. Many interesting mathematical statements cover an infinite number of special cases, but we are not able to test them for an infinite number of cases. Incomplete induction often leads to erroneous results.

In many cases, the way out of this kind of difficulty is to resort to a special method of reasoning, called the method of mathematical induction. It is as follows.

Suppose you need to prove the validity of a certain statement for any natural number n (for example, you need to prove that the sum of the first n odd numbers is equal to n 2). Direct verification of this statement for each value of n is impossible, since the set of natural numbers is infinite. To prove this statement, first check its validity for n=1. Then they prove that for any natural value of k, the validity of the statement under consideration for n=k implies its validity for n=k+1.

Then the statement is considered proven for all n. In fact, the statement is true for n=1. But then it is also true for the next number n=1+1=2. The validity of the statement for n=2 implies its validity for n=2+

1=3. This implies the validity of the statement for n=4, etc. It is clear that, in the end, we will reach any natural number n. This means that the statement is true for any n.

Summarizing what has been said, we formulate the following general principle.

The principle of mathematical induction.

If proposal A(n), depending on the natural numbern, true forn=1 and from the fact that it is true forn=k(Wherek-any natural number), it follows that it is true for the next numbern=k+1, then assumption A(n) true for any natural numbern.

In a number of cases, it may be necessary to prove the validity of a certain statement not for all natural numbers, but only for n>p, where p is a fixed natural number. In this case, the principle of mathematical induction is formulated as follows. If proposal A(n) true forn=pand if A(k) Þ A(k+1)for anyonek>p,then sentence A(n)true for anyonen>p.

The proof using the method of mathematical induction is carried out as follows. First, the statement to be proved is checked for n=1, i.e. the truth of statement A(1) is established. This part of the proof is called the induction basis. Then comes the part of the proof called the induction step. In this part, they prove the validity of the statement for n=k+1 under the assumption of the validity of the statement for n=k (induction assumption), i.e. prove that A(k)ÞA(k+1).

EXAMPLE 1

Prove that 1+3+5+…+(2n-1)=n 2.

Solution: 1) We have n=1=1 2 . Hence,

the statement is true for n=1, i.e. A(1) is true.

2) Let us prove that A(k)ÞA(k+1).

Let k be any natural number and let the statement be true for n=k, i.e.

1+3+5+…+(2k-1)=k 2 .

Let us prove that then the statement is also true for the next natural number n=k+1, i.e. What

1+3+5+…+(2k+1)=(k+1) 2 .

Indeed,

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

So, A(k)ÞA(k+1). Based on the principle of mathematical induction, we conclude that assumption A(n) is true for any nÎN.

EXAMPLE 2

Prove that

1+x+x 2 +x 3 +…+x n =(x n+1 -1)/(x-1), where x¹1

Solution: 1) For n=1 we get

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

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

2) Let k be any natural number and let the formula be true for n=k, i.e.

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

Let us prove that then the equality

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

Indeed

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).

So, A(k)ÞA(k+1). Based on the principle of mathematical induction, we conclude that the formula is true for any natural number n.

EXAMPLE 3

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

Solution: 1) For n=3 the statement is true

And 3 is meaningful, because in a triangle

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

A 2 A(3) is true.

2) Let us assume that in every

a convex k-gon has-

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

And k Let us prove that then in a convex

(k+1)-gon number

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

Let A 1 A 2 A 3 …A k A k+1 be a convex (k+1)-gon. Let's draw a diagonal A 1 A k in it. To calculate the total number of diagonals of this (k+1)-gon, you need to count the number of diagonals in the k-gon A 1 A 2 ...A k , add k-2 to the resulting number, i.e. the number of diagonals of the (k+1)-gon emanating from the vertex A k+1, and, in addition, the diagonal A 1 A k should be taken into account.

Thus,

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

So, A(k)ÞA(k+1). Due to the principle of mathematical induction, the statement is true for any convex n-gon.

EXAMPLE 4

Prove that for any n the following statement is true:

1 2 +2 2 +3 2 +…+n 2 =n(n+1)(2n+1)/6.

Solution: 1) Let n=1, then

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

This means that for n=1 the statement is true.

2) Assume that n=k

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

3) Consider this statement 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 proven the equality to be true for n=k+1, therefore, by virtue of the method of mathematical induction, the statement is true for any natural number n.

EXAMPLE 5

Prove that for any natural number n the equality is true:

1 3 +2 3 +3 3 +…+n 3 =n 2 (n+1) 2 /4.

Solution: 1) Let n=1.

Then X 1 =1 3 =1 2 (1+1) 2 /4=1.

We see that for n=1 the statement is true.

2) Suppose that the equality is true for n=k

If a sentence A(n), depending on a natural number n, is true for n=1 and from the fact that it is true for n=k (where k is any natural number), it follows that it is also true for the next number n=k +1, then assumption A(n) is true for any natural number n.

In a number of cases, it may be necessary to prove the validity of a certain statement not for all natural numbers, but only for n>p, where p is a fixed natural number. In this case, the principle of mathematical induction is formulated as follows.

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

The proof using the method of mathematical induction is carried out as follows. First, the statement to be proved is checked for n=1, i.e. the truth of statement A(1) is established. This part of the proof is called the induction basis. Then comes the part of the proof called the induction step. In this part, they prove the validity of the statement for n=k+1 under the assumption of the validity of the statement for n=k (induction assumption), i.e. prove that A(k) 1 A(k+1)

Prove that 1+3+5+…+(2n-1)=n 2.

• 1) We have n=1=1 2 . Therefore, the statement is true for n=1, i.e. A(1) true
• 2) Let us prove that A(k) ≥ A(k+1)

Let k be any natural number and let the statement be true for n=k, i.e.

1+3+5+…+(2k-1)=k 2

Let us prove that then the statement is also true for the next natural number n=k+1, i.e. What

• 1+3+5+…+(2k+1)=(k+1) 2 Indeed,
• 1+3+5+…+(2k-1)+(2k+1)=k 2 +2k+1=(k+1) 2

So, A(k) 1 A(k+1). Based on the principle of mathematical induction, we conclude that assumption A(n) is true for any n O N

Prove that

1+x+x 2 +x 3 +…+x n =(x n+1 -1)/(x-1), where x No. 1

• 1) For n=1 we get
• 1+x=(x 2 -1)/(x-1)=(x-1)(x+1)/(x-1)=x+1

therefore, for n=1 the formula is correct; A(1) true

• 2) Let k be any natural number and let the formula be true for n=k,
• 1+x+x 2 +x 3 +…+x k =(x k+1 -1)/(x-1)

Let us prove that then the equality

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

So, A(k) 1 A(k+1). Based on the principle of mathematical induction, we conclude that the formula is true for any natural number n

Prove that the number of diagonals of a convex n-gon is n(n-3)/2

Solution: 1) For n=3 the statement is true, because in the triangle

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

2) Suppose that in every convex k-gon there are A 1 x A k =k(k-3)/2 diagonals. A k Let us prove that then in a convex A k+1 (k+1)-gon the number of diagonals A k+1 =(k+1)(k-2)/2.

Let A 1 A 2 A 3 …A k A k+1 be a convex (k+1)-gon. Let's draw a diagonal A 1 A k in it. To calculate the total number of diagonals of this (k+1)-gon, you need to count the number of diagonals in the k-gon A 1 A 2 ...A k , add k-2 to the resulting number, i.e. the number of diagonals of the (k+1)-gon emanating from the vertex A k+1, and, in addition, the diagonal A 1 A k should be taken into account

Thus,

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

So, A(k) 1 A(k+1). Due to the principle of mathematical induction, the statement is true for any convex n-gon.

Prove that for any n the following statement is true:

1 2 +2 2 +3 2 +…+n 2 =n(n+1)(2n+1)/6

Solution: 1) Let n=1, then

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

2) Assume that n=k

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

3) Consider this statement 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 proven the equality to be true for n=k+1, therefore, by virtue of the method of mathematical induction, the statement is true for any natural number n

Prove that for any natural number n the equality is true:

1 3 +2 3 +3 3 +…+n 3 =n 2 (n+1) 2 /4

Solution: 1) Let n=1

Then X 1 =1 3 =1 2 (1+1) 2 /4=1. We see that for n=1 the statement is true.

2) Suppose that the equality is true for n=k

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

3) Let us prove the truth of this statement for n=k+1, i.e.

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 statement is true for n=k+1, therefore, the equality is true for any natural number n

Prove that

((2 3 +1)/(2 3 -1)) ґ ((3 3 +1)/(3 3 -1)) ґ ... ґ ((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 ґ 2 ґ 3)/2(2 2 +2+1), i.e. it's true
• 2) Assume that the expression is true for n=k
• (2 3 +1)/(2 3 -1) ґ … ґ (k 3 +1)/(k 3 -1)=3k(k+1)/2(k 2 +k+1)
• 3) Let us prove the validity of the expression for n=k+1
• (((2 3 +1)/(2 3 -1)) ґ … ґ ((k 3 +1)/(k 3 -1))) ґ (((k+1) 3 +

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

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

ґ ((k+1) 2 +(k+1)+1)

We have proven the equality to be true for n=k+1, therefore, by virtue of the method of mathematical induction, the statement is true for any n>2

Prove that

1 3 -2 3 +3 3 -4 3 +…+(2n-1) 3 -(2n) 3 =-n 2 (4n+3) for any natural number n

Solution: 1) Let n=1, then

• 1 3 -2 3 =-1 3 (4+3); -7=-7
• 2) Suppose 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 statement for 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)

The validity of the equality for n=k+1 has also been proven, therefore the statement is true for any natural number n.

Prove the identity is correct

(1 2 /1 ґ 3)+(2 2 /3 ґ 5)+…+(n 2 /(2n-1) ґ (2n+1))=n(n+1)/2(2n+1) for any natural n

• 1) For n=1 the identity is true 1 2 /1 ґ 3=1(1+1)/2(2+1)
• 2) Suppose that for n=k
• (1 2 /1 ґ 3)+…+(k 2 /(2k-1) ґ (2k+1))=k(k+1)/2(2k+1)
• 3) Let us prove that the identity is true for n=k+1
• (1 2 /1 ґ 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)) ґ ((k/2 )+((k+1)/(2k+3)))=(k+1)(k+2) ґ (2k+1)/2(2k+1)(2k+3)=(k+1 )(k+2)/2(2(k+1)+1)

From the above proof it is clear that the statement is true for any natural number n.

Prove that (11 n+2 +12 2n+1) is divisible by 133 without remainder

Solution: 1) Let n=1, then

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

But (23 ґ 133) is divisible by 133 without a remainder, which means that for n=1 the statement is true; A(1) is true.

• 2) Suppose that (11 k+2 +12 2k+1) is divisible by 133 without a remainder
• 3) Let us prove that in this case (11 k+3 +12 2k+3) is divisible by 133 without a remainder. Indeed
• 11 k+3 +12 2l+3 =11 ґ 11 k+2 +12 2 ґ 12 2k+1 =11 ґ 11 k+2 +

+(11+133) ґ 12 2k+1 =11(11 k+2 +12 2k+1)+133 ґ 12 2k+1

The resulting sum is divided by 133 without a remainder, since its first term is divisible by 133 without a remainder by assumption, and in the second one of the factors is 133. So, A(k) 1 A(k+1). By virtue of the method of mathematical induction, the statement is proven

Prove that for any n 7 n -1 is divisible by 6 without a remainder

• 1) Let n=1, then X 1 =7 1 -1=6 is divided by 6 without a remainder. This means that for n=1 the statement is true
• 2) Suppose that when n=k 7 k -1 is divided by 6 without a remainder
• 3) Let us prove that the statement is true for n=k+1

X k+1 =7 k+1 -1=7 ґ 7 k -7+6=7(7 k -1)+6

The first term is divisible by 6, since 7 k -1 is divisible by 6 by assumption, and the second term is 6. This means 7 n -1 is a multiple of 6 for any natural number n. By virtue of the method of mathematical induction, the statement is proven.

Prove that 3 3n-1 +2 4n-3 for an arbitrary natural number n is divisible by 11.

1) Let n=1, then

X 1 =3 3-1 +2 4-3 =3 2 +2 1 =11 is divided by 11 without a remainder.

This means that for n=1 the statement is true

• 2) Suppose that when n=k X k =3 3k-1 +2 4k-3 is divided by 11 without a remainder
• 3) Let us prove that the statement is true 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 ґ 3 3k-1 +2 4 ґ 2 4k-3 =

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

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

The first term is divisible by 11 without a remainder, since 3 3k-1 +2 4k-3 is divisible by 11 by assumption, the second is divisible by 11, because one of its factors is the number 11. This means that the sum is divisible by 11 without a remainder for any natural number n. By virtue of the method of mathematical induction, the statement is proven.

Prove that 11 2n -1 for an arbitrary natural number n is divisible by 6 without a remainder

• 1) Let n=1, then 11 2 -1=120 is divisible by 6 without a remainder. This means that for n=1 the statement is true
• 2) Suppose that when n=k 1 2k -1 is divided by 6 without a remainder
• 11 2(k+1) -1=121 ґ 11 2k -1=120 ґ 11 2k +(11 2k -1)

Both terms are divisible by 6 without a remainder: the first contains a multiple of 6, 120, and the second is divisible by 6 without a remainder by assumption. This means that the sum is divisible by 6 without a remainder. By virtue of the method of mathematical induction, the statement is proven.

Prove that 3 3n+3 -26n-27 for an arbitrary natural number n is divisible by 26 2 (676) without a remainder

Let us first prove that 3 3n+3 -1 is divisible by 26 without a remainder

• 1. When n=0
• 3 3 -1=26 is divided by 26
• 2. Suppose that for n=k
• 3 3k+3 -1 is divisible by 26
• 3. Let us prove that the statement is true for n=k+1
• 3 3k+6 -1=27 ґ 3 3k+3 -1=26 ґ 3 3л+3 +(3 3k+3 -1) -divided by 26

Now let's prove the statement formulated in the problem statement

• 1) Obviously, for n=1 the statement is true
• 3 3+3 -26-27=676
• 2) Suppose that for n=k the expression 3 3k+3 -26k-27 is divided by 26 2 without a remainder
• 3) Let us prove that the statement is true for n=k+1
• 3 3k+6 -26(k+1)-27=26(3 3k+3 -1)+(3 3k+3 -26k-27)

Both terms are divisible by 26 2; the first is divisible by 26 2 because we have proven the expression in parentheses is divisible by 26, and the second is divisible by the induction hypothesis. By virtue of the method of mathematical induction, the statement is proven

Prove that if n>2 and x>0, then the inequality (1+x) n >1+n ґ x is true

• 1) For n=2 the inequality is valid, since
• (1+x) 2 =1+2x+x 2 >1+2x

So A(2) is true

• 2) Let us prove that A(k) ≈ A(k+1), if k> 2. Assume that A(k) is true, i.e., that the inequality
• (1+x) k >1+k ґ x. (3)

Let us prove that then A(k+1) is also true, i.e., that the inequality

(1+x) k+1 >1+(k+1) ґ x

In fact, multiplying both sides of inequality (3) by the positive number 1+x, we obtain

(1+x) k+1 >(1+k ґ x)(1+x)

Consider the right side of the last inequality; we have

(1+k ґ x)(1+x)=1+(k+1) ґ x+k ґ x 2 >1+(k+1) ґ x

As a result, we get that (1+x) k+1 >1+(k+1) ґ x

So, A(k) 1 A(k+1). Based on the principle of mathematical induction, it can be argued that Bernoulli’s inequality is valid for any n> 2

Prove that the inequality (1+a+a 2) m > 1+m ґ a+(m(m+1)/2) ґ a 2 for a> 0 is true

Solution: 1) When m=1

• (1+a+a 2) 1 > 1+a+(2/2) ґ a 2 both sides are equal
• 2) Suppose that for m=k
• (1+a+a 2) k >1+k ґ a+(k(k+1)/2) ґ a 2
• 3) Let us prove that for m=k+1 the inequality is true
• (1+a+a 2) k+1 =(1+a+a 2)(1+a+a 2) k >(1+a+a 2)(1+k ґ a+

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

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

+((k+1)(k+2)/2) ґ a 2

We have proven the inequality to be true for m=k+1, therefore, by virtue of the method of mathematical induction, the inequality is valid for any natural number m

Prove that for n>6 the inequality 3 n >n ґ 2 n+1 is true

Let us rewrite the inequality in the form (3/2) n >2n

• 1. For n=7 we have 3 7 /2 7 =2187/128>14=2 ґ 7 the inequality is true
• 2. Suppose that for n=k (3/2) k >2k
• 3) Let us prove the inequality for n=k+1
• 3 k+1 /2 k+1 =(3 k /2 k) ґ (3/2)>2k ґ (3/2)=3k>2(k+1)

Since k>7, the last inequality is obvious.

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

Prove that for n>2 the inequality is true

1+(1/2 2)+(1/3 2)+…+(1/n 2)<1,7-(1/n)

• 1) For n=3 the inequality is true
• 1+(1/2 2)+(1/3 2)=245/180
• 2. Suppose that for n=k
• 1+(1/2 2)+(1/3 2)+…+(1/k 2)=1.7-(1/k)
• 3) Let us prove the validity of the inequality for n=k+1
• (1+(1/2 2)+…+(1/k 2))+(1/(k+1) 2)

Let us prove that 1.7-(1/k)+(1/(k+1) 2)<1,7-(1/k+1) Ы

S (1/(k+1) 2)+(1/k+1)<1/k Ы (k+2)/(k+1) 2 <1/k Ы

ы k(k+2)<(k+1) 2 Ы k 2 +2k

The latter is obvious, and therefore

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

By virtue of the method of mathematical induction, the inequality is proven.

In many branches of mathematics it is necessary to prove the truth of a statement depending on , i.e. truth of statement p(n) For " n ON (for any n ON p(n) right).

This can often be proven by the method of mathematical induction.

This method is based on the principle of mathematical induction. It is usually chosen as one of the axioms of arithmetic and is therefore accepted without proof. According to the principle of mathematical induction, the sentence p(n) is considered true for all natural values ​​of the variable if two conditions are met:

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

2. From the sentence that p(n) true for n =k (k — arbitrary natural number) it follows that it is true for n =k+ 1.

The method of mathematical induction means the following method of proof

1. Check the truth of the statement for n= 1 – base of induction.

2. Assume that the statement is true for n = k – inductive hypothesis.

3. They prove that then it is also true for n =k+ 1 inductive junction.

Sometimes a suggestion p(n) turns out to be true not for all naturals n, and starting from some for n = n 0. In this case, the truth of p(n) at n = n 0.

Example 1. Let . Prove that

1. Induction base: at n= 1 by definition S 1 = 1 and according to the formula we get one result. The statement is true.

n = k And .

n = k+ 1. Let us prove that .

Indeed, by virtue of the inductive assumption

Let's transform this expression

Inductive transition has been proven.

Comment. It is useful to write down what is given (the inductive hypothesis) and what needs to be proven!

Example 2. Prove

1. Base of induction. At n= 1, the statement is obviously true.

2. Inductive hypothesis. Let n = k And

3. Inductive transition. Let n = k+ 1. Let us prove:

Indeed, let's square the right side as the sum of two numbers:

Using the inductive assumption and the formula for the sum of an arithmetic progression: , we obtain

Example 3. Prove inequality

1. The basis of induction in this case is checking the truth of the statement for , i.e. it is necessary to check the inequality. To do this, it is enough to square the inequality: or 63< 64 – неравенство верно.

2. Let the inequality be true for , i.e.

3. Let us prove:

We use the induction assumption

Knowing what the right side in the inequality being proved should look like, let’s highlight this part

It remains to establish that the extra factor does not exceed one. Really,

Example 4. Prove that for any natural number the number ends in the digit .

1. The smallest natural number from which the statement is valid is equal to . .

2. Let at number end in . This means that this number can be written in the form , where is some natural number. Then .

3. Let . Let's prove that it ends in . Using the obtained representation, we obtain

The last number has exactly ones.

Application

1.4. Method of mathematical induction

As you know, mathematical statements (theorems) must be substantiated and proven. We will now get acquainted with one of the methods of proof - the method of mathematical induction.

In a broad sense, induction is a method of reasoning that allows one to move from particular statements to general ones. The reverse transition, from general statements to specific ones, is called deduction.

Deduction always leads to correct conclusions. For example, we know the general result: all integers ending in zero are divisible by 5. From this, of course, we can conclude that any specific number ending in 0, for example 180, is divisible by 5.

At the same time, induction can lead to incorrect conclusions. For example, noticing that the number 60 is divisible by the numbers 1, 2, 3, 4, 5, 6, we have no right to conclude that 60 is divisible by any number at all.

The method of mathematical induction allows in many cases to strictly prove the validity of the general statement P(n), the formulation of which includes the natural number n.

Application of the method includes 3 stages.

1) Base of induction: we check the validity of the statement P(n) for n = 1 (or for another, particular value of n, starting from which the validity of P(n) is assumed).

2) Induction assumption: we assume that P(n) is valid for n = k.

3) Induction step: using the assumption, we prove that P(n) is valid for n = k + 1.

As a result, we can conclude that P(n) is valid for any n ∈ N. Indeed, for n = 1 the statement is true (base of induction). And therefore, it is also true for n = 2, since the transition from n = 1 to n = 2 is justified (induction step). Applying the induction step again and again, we obtain the validity of P(n) for n = 3, 4, 5, . . ., i.e., the validity of P(n) for all n.

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

+ (2n - 1) = n2.

We will carry out the proof using the method of mathematical induction.

1) Base: with n=1 there is only one term on the left, we get: 1 = 1.

The statement is true.

2) Assumption: we assume that for some k the equality is true: 1 + 3 + 5 + … + (2k - 1) = k2.

Solving problems about the probability of hits during shots

The general formulation of the problem is as follows:

The probability of hitting the target with one shot is $p$. $n$ shots are fired. Find the probability that the target will be hit exactly $k$ times (there will be $k$ hits).

We apply Bernoulli's formula and get:

$$ 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 $C_n^k$ is the number of combinations of $n$ by $k$.

If the problem involves several arrows with different probabilities hitting the target, theory, example solutions and a calculator can be found here.

Video tutorial and Excel template

Watch our video on solving Bernoulli shot problems and learn how to use Excel to solve common problems.

The Excel calculation file from the video can be downloaded for free and used to solve your problems.

Examples of solutions to problems about hitting a target in a series of shots

Let's look at a few typical examples.

Example 1. Fired 7 shots. The probability of a hit with one shot is 0.705. Find the probability that there will be exactly 5 hits.

We get that the problem involves repeated independent tests (shots at a target), a total of $n=7$ shots are fired, the probability of a hit for each $p=0.705$, the probability of a miss $q=1-p=1-0.705=0.295 $.

We need to find that there will be exactly $k=5$ hits. We substitute everything into formula (1) and get: $$ P_7(5)=C_(7)^5 \cdot 0.705^5 \cdot 0.295^2 = 21\cdot 0.705^5 \cdot 0.295^2= 0.318. $$

Example 2. The probability of hitting the target with one shot is 0.4.

Four independent shots are fired at the target. Find the probability that there will be at least one hit on the target.

We study the problem and write down the parameters: $n=4$ (shot), $p=0.4$ (probability of a hit), $k \ge 1$ (there will be at least one hit).

We use the formula for the probability of the opposite event (there is not a single hit):

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

The probability of hitting at least one time out of four is 0.87 or 87%.

Example 3. The probability of hitting the target by the shooter is 0.3.

Find the probability that with 6 shots the target will be hit from three to six times.

Unlike previous problems, here you need to find the probability that the number of hits will be in a certain interval (and not exactly equal to some number). But the same formula is used.

Let's find the probability that the target will be hit from three to six times, that is, there will be either 3, or 4, or 5, or 6 hits.

We calculate these probabilities using 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.

Since the events are incompatible, the desired probability can be found using the formula for adding probabilities: $$ P_6(3 \le k \le 6)=P_6(3)+P_6(4)+P_6(5)+P_6(6)=$$ $$ = 0.185+0.06+0.01+0.001=0.256.$$

Example 4. The probability of at least one hit on the target with four shots is 0.9984. Find the probability of hitting the target with one shot.

Let us denote the probability of hitting the target with one shot. Let's introduce an event:
$A = $ (Out of four shots at least one will hit the target),
as well as the opposite event, which can be written as:
$\overline(A) = $ (All 4 shots will miss the target, not a single hit).

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

Let's write down the known values: $n=4$, $P(A)=0.9984$. Substitute into formula (1) and get:

$$ 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 solve the resulting equation:

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

So, the probability of hitting the target with one shot is 0.8.

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Solving inequalities using a calculator

Inequalities in mathematics refer to all equations where "=" is replaced by any of the following symbols: \[>\]\[\geq\]\[

* linear;

* square;

* fractional;

* indicative;

* trigonometric;

* logarithmic.

Depending on this, inequalities are called linear, partial, etc.

You should be aware of these signs:

* inequalities with greater than (>) or less than (

* Inequalities with symbols that are greater than or equal to \[\geq\] less than or equal to [\leq\] are called unprofessional;

* the icon is not the same \[\ne\] one, but it is necessary to resolve cases with this icon all the time.

Such inequality is solved through identity transformations.

Also read our article Solve Complete Solution for Online Equation

Let us assume that the following inequality holds:

We solve it in the same way as a linear equation, but we need to be careful about the inequality sign.

First we move terms from the unknown to the left, from the known to the right, reversing the symbols:

Then we divide both sides by -4 and reverse the inequality sign:

This is the answer to this equation.

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You can solve the equation on our website pocketteacher.ru.

Bernoulli inequality calculator

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Method of complete mathematical induction

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The proof method based on Peano's axiom 4 is used to prove many mathematical properties and various statements. The basis for this is the following theorem.

Theorem. If the statement A(n) with natural variable n true for n= 1 and from the fact that it is true for n = k, it follows that it is true for the next number n=k, then the statement A(n) n.

Proof. Let us denote by M the set of those and only those natural numbers for which the statement A(n) true. Then from the conditions of the theorem we have: 1) 1 M; 2) k MkM. From here, based on axiom 4, we conclude that M =N, i.e. statement A(n) true for any natural n.

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

1) prove that the statement A(n) true for n= A(1);

2) assume that the statement A(n) true for n = k, and, based on this assumption, prove that the statement A(n) true for n = k + 1, i.e. that the statement is true A(k) A(k + 1).

If A( 1) A(k) A(k + 1) - true statement, then they conclude that the statement A(n) true for any natural number n.

Proof by the method of mathematical induction can begin not only with confirmation of the truth of the statement for n= 1, but also from any natural number m. In this case the statement A(n) will be proven for all natural numbers nm.

Problem: Let us prove that for any natural number the equality 1 + 3 + 5 … + (2 n- 1) = n.

Solution. Equality 1 + 3 + 5 … + (2 n- 1) = n is a formula that can be used to find the sum of the first consecutive odd natural numbers. For example, 1 + 3 + 5 + 7 = 4= 16 (sum contains 4 terms), 1 + 3 + 5 + 7 + 9 + 11 = 6= 36 (sum contains 6 terms); if this sum contains 20 terms of the indicated type, then it is equal to 20 = 400, etc. Having proven the truth of this equality, we will be able to find the sum of any number of terms of the specified type using the formula.

1) Let us verify the truth of this equality for n= 1. When n= 1 the left side of the equality consists of one term equal to 1, the right side is equal to 1= 1. Since 1 = 1, then for n= 1 this equality is true.

2) Suppose that this equality is true for n = k, i.e. that 1 + 3 + 5 + … + (2 k- 1) = k. Based on this assumption, we prove that it is true for n = k + 1, i.e. 1 + 3 + 5 + … + (2 k- 1) + (2(k + 1) - 1) = (k + 1).

Let's look at the left side of the last equality.

By assumption, the sum of the first k terms is equal to k and therefore 1 + 3 + 5 + … + (2 k- 1) + (2(k + 1) - 1) = 1 + 3 + 5 + … + (2k- 1) + (2k+ 1)=

= k+(2k + 1) = k+ 2k + 1. Expression k+ 2k + 1 is identically equal to the expression ( k + 1).

Therefore, the truth of this equality for n = k + 1 has been proven.

Thus, this equality is true for n= 1 and from its truth for n = k must be true for n = k + 1.

This proves that this equality is true for any natural number.

Using the method of mathematical induction, you can prove the truth of not only equalities, but also inequalities.

Task. Prove that , where nN.

Solution. Let us check the truth of the inequality at n= 1. We have - true inequality.

Let us assume that the inequality is true for n = k, those. - true inequality. Let us prove, based on the assumption, that it is also true for n = k + 1, i.e. (*).

Let's transform the left side of the inequality (*), taking into account that: .

But that means .

So, this inequality is true for n= 1, and, from the fact that the inequality is true for some n= k, we found that it is also true for n= k + 1.

Thus, using axiom 4, we proved that this inequality is true for any natural number.

Other statements can be proven using the method of mathematical induction.

Task. Prove that for any natural number the statement is true.

Solution. Let's check the truth of the statement when n= 1: -true statement.

Let us assume that this statement is true for n = k: . Let us show, using this, the truth of the statement when n = k + 1: .

Let's transform the expression: . Let's find the difference k And k+ 1 members. If it turns out that the resulting difference is a multiple of 7, and by assumption the subtrahend is divisible by 7, then the minuend is also a multiple of 7:

The product is a multiple of 7, therefore, and .

Thus, this statement is true for n= 1 and from its truth for n = k must be true for n = k + 1.

This proves that this statement is true for any natural number.

Task. Prove that for any natural number n 2 statement (7-1)24 is true.

Solution. 1) Let's check the truth of the statement when n= 2: - true statement.