NCERT Solutions For Class 11 Maths Chapter 4 Principle of Mathematical Induction are given in an easy way at BYJUâ€™S. Students learn about the Principle of Mathematical Induction and its application in detail through this chapter. After solving all the questions present in the NCERT textbook and Exemplar, students can easily score maximum marks for the questions of this chapter.Â

Principle of Mathematical Induction is a specific technique used to prove certain mathematically accepted statements in algebra and in other applications of Mathematics, such as inductive and deductive reasoning. NCERT Solutions of BYJUâ€™S cover all these concepts and help in scoring full marks for this chapter. These solutions are useful for further studies and for those who are preparing for competitive exams. NCERT Solutions For Class 11 Maths are very accurate and make it easy to crack the exam with good marks.

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Exercise 4.1 Solutions 24 Questions

### Access NCERT Solutions for Class 11 Maths Chapter 4

Exercise 4.1 page: 94

**Prove the following by using the principle of mathematical induction for all n âˆˆ N:**

**1. **

**Solution:**

P (k + 1) is true whenever P (k) is true.

Therefore, by the principle of mathematical induction, statement P (n) is true for all natural numbers i.e. n.

**2. **

**Solution:**

P (k + 1) is true whenever P (k) is true.

Therefore, by the principle of mathematical induction, statement P (n) is true for all natural numbers i.e. n.

**3.**

**Solution:**

P (k + 1) is true whenever P (k) is true.

Therefore, by the principle of mathematical induction, statement P (n) is true for all natural numbers i.e. n.

**4.**

**Solution:**

P (k + 1) is true whenever P (k) is true.

**5.**

**Solution:**

P (k + 1) is true whenever P (k) is true.

**6.**

**Solution:**

P (k + 1) is true whenever P (k) is true.

**7.**

**Solution:**

P (k + 1) is true whenever P (k) is true.

**8. 1.2 + 2.2 ^{2}Â + 3.2^{2}Â + â€¦ +Â n.2^{n}Â = (nÂ â€“ 1) 2^{n}^{+1}Â + 2**

**Solution:**

We can write the given statement as

P (*n*): 1.2 + 2.2^{2}Â + 3.2^{2}Â + â€¦ +Â *n*.2* ^{n}*Â = (

*n*Â â€“ 1) 2

^{n}^{+1}Â + 2

If n = 1 we get

P (1): 1.2 = 2 = (1 â€“ 1) 2^{1+1}Â + 2 = 0 + 2 = 2

Which is true.

Consider P (k) be true for some positive integer k

1.2 + 2.2^{2}Â + 3.2^{2}Â + â€¦ +Â *k.*2* ^{k}*Â = (

*k*Â â€“ 1) 2

^{k}^{Â + 1}Â + 2 â€¦ (i)

Now let us prove that P (k + 1) is true.

Here

P (k + 1) is true whenever P (k) is true.

**9.**

**Solution:**

We can write the given statement as

P (k + 1) is true whenever P (k) is true.

**10.**

**Solution:**

P (k + 1) is true whenever P (k) is true.

**11.**

**Solution:**

P (k + 1) is true whenever P (k) is true.

**12.**

**Solution:**

P (k + 1) is true whenever P (k) is true.

**13.**

**Solution:**

P (k + 1) is true whenever P (k) is true.

**14.**

**Solution:**

By further simplification

= (k + 1) + 1

P (k + 1) is true whenever P (k) is true.

**15.**

**Solution:**

P (k + 1) is true whenever P (k) is true.

**16. **

**Solution:**

P (k + 1) is true whenever P (k) is true.

**17.**

**Solution:**

P (k + 1) is true whenever P (k) is true.

**18. **

**Solution:**

We can write the given statement as

P (k + 1) is true whenever P (k) is true.

**19. nÂ (nÂ + 1) (nÂ + 5) is a multiple of 3**

**Solution:**

We can write the given statement as

P (n): n (n + 1) (n + 5), which is a multiple of 3

If n = 1 we get

1 (1 + 1) (1 + 5) = 12, which is a multiple of 3

Which is true.

Consider P (k) be true for some positive integer k

k (k + 1) (k + 5) is a multiple of 3

k (k + 1) (k + 5) = 3m, where m âˆˆÂ N â€¦â€¦ (1)

Now let us prove that P (k + 1) is true.

Here

(k + 1) {(k + 1) + 1} {(k + 1) + 5}

We can write it as

= (k + 1) (k + 2) {(k + 5) + 1}

By multiplying the terms

= (k + 1) (k + 2) (k + 5) + (k + 1) (k + 2)

So we get

= {k (k + 1) (k + 5) + 2 (k + 1) (k + 5)} + (k + 1) (k + 2)

Substituting equation (1)

= 3m + (k + 1) {2 (k + 5) + (k + 2)}

By multiplication

= 3m + (k + 1) {2k + 10 + k + 2}

On further calculation

= 3m + (k + 1) (3k + 12)

Taking 3 as common

= 3m + 3 (k + 1) (k + 4)

We get

= 3 {m + (k + 1) (k + 4)}

= 3 Ã— q where q = {m + (k + 1) (k + 4)} is some natural number

(k + 1) {(k + 1) + 1} {(k + 1) + 5} is a multiple of 3

P (k + 1) is true whenever P (k) is true.

**20. 10 ^{2}^{n}^{Â â€“ 1Â }+ 1 is divisible by 11**

**Solution:**

We can write the given statement as

P (*n*): 10^{2}^{n}^{Â â€“ 1Â }+ 1 is divisible by 11

If n = 1 we get

P (1) = 10^{2.1 â€“ 1Â }+ 1 = 11, which is divisible by 11

Which is true.

Consider P (k) be true for some positive integer k

10^{2}^{k}^{Â â€“ 1Â }+ 1 is divisible by 11

10^{2}^{k}^{Â â€“ 1Â }+ 1 = 11*m*, whereÂ *m*Â âˆˆÂ N**Â **â€¦â€¦ (1)

Now let us prove that P (k + 1) is true.

Here

10 ^{2 (k + 1) â€“ 1} + 1

We can write it as

= 10 ^{2k + 2 â€“ 1} + 1

= 10 ^{2k + 1} + 1

By addition and subtraction of 1

= 10 ^{2} (10^{2k-1} + 1 â€“ 1) + 1

We get

= 10 ^{2} (10^{2k-1} + 1) â€“ 10^{2} + 1

Using equation 1 we get

= 10^{2}. 11m â€“ 100 + 1

= 100 Ã— 11m â€“ 99

Taking out the common terms

= 11 (100m â€“ 9)

= 11 r, where r = (100m â€“ 9) is some natural number

10 ^{2(k + 1) â€“ 1 }+ 1 is divisible by 11

P (k + 1) is true whenever P (k) is true.

**21. x^{2}^{n}Â â€“Â y^{2}^{n}Â is divisible byÂ xÂ +Â y**

**Solution:**

We can write the given statement as

P (*n*):Â *x*^{2}* ^{n}*Â â€“Â

*y*

^{2}

*Â is divisible by*

^{n}*Â xÂ*+Â

*y*

If n = 1 we get

P (1) = *x*^{2Â Ã—Â 1}Â â€“Â *y*^{2Â Ã—Â 1}Â =Â *x*^{2}Â â€“Â *y*^{2}Â = (*xÂ *+Â *y*) (*x*Â â€“Â *y*), which is divisible by (x + y)

Which is true.

Consider P (k) be true for some positive integer k

*x*^{2}* ^{k}*Â â€“Â

*y*

^{2}

*Â is divisible by*

^{k}*Â xÂ*+Â

*y*

*x*^{2}* ^{k}*Â â€“Â

*y*

^{2}

*Â =Â*

^{k}*m*Â (

*xÂ*+Â

*y*), whereÂ

*m*Â âˆˆÂ N

**Â**â€¦â€¦ (1)

Now let us prove that P (k + 1) is true.

Here

x ^{2(k + 1)} â€“ y ^{2(k + 1)}

We can write it as

= x ^{2k} . x^{2} â€“ y^{2k} . y^{2}

By adding and subtracting y^{2k} we get

= x^{2} (x^{2k} â€“ y^{2k} + y^{2k}) â€“ y^{2k}. y^{2}

From equation (1) we get

= x^{2} {m (x + y) + y^{2k}} â€“ y^{2k}. y^{2}

By multiplying the terms

= m (x + y) x^{2} + y^{2k}. x^{2} â€“ y^{2k}. y^{2}

Taking out the common terms

= m (x + y) x^{2} + y^{2k} (x^{2} â€“ y^{2})

Expanding using formula

= m (x + y) x^{2} + y^{2k} (x + y) (x â€“ y)

So we get

= (x + y) {mx^{2} + y^{2k} (x â€“ y)}, which is a factor of (x + y)

P (k + 1) is true whenever P (k) is true.

**22. 3 ^{2}^{n}^{Â + 2}Â â€“ 8nÂ â€“ 9 is divisible by 8**

**Solution:**

We can write the given statement as

P (*n*): 3^{2}^{n}^{Â + 2}Â â€“ 8*n*Â â€“ 9 is divisible by 8

If n = 1 we get

P (1) = 3^{2Â Ã—Â 1 + 2}Â â€“ 8 Ã— 1 â€“ 9 = 64, which is divisible by 8

Which is true.

Consider P (k) be true for some positive integer k

3^{2}^{k}^{Â + 2}Â â€“ 8*k*Â â€“ 9 is divisible by 8

3^{2}^{k}^{Â + 2}Â â€“ 8*k*Â â€“ 9 = 8*m*, whereÂ *m*Â âˆˆÂ N**Â **â€¦â€¦ (1)

Now let us prove that P (k + 1) is true.

Here

3 ^{2(k + 1) + 2} â€“ 8 (k + 1) â€“ 9

We can write it as

= 3 ^{2k + 2} . 3^{2} â€“ 8k â€“ 8 â€“ 9

By adding and subtracting 8k and 9 we get

= 3^{2} (3^{2k + 2} â€“ 8k â€“ 9 + 8k + 9) â€“ 8k â€“ 17

On further simplification

= 3^{2} (3^{2k + 2} â€“ 8k â€“ 9) + 3^{2} (8k + 9) â€“ 8k â€“ 17

From equation (1) we get

= 9. 8m + 9 (8k + 9) â€“ 8k â€“ 17

By multiplying the terms

= 9. 8m + 72k + 81 â€“ 8k â€“ 17

So we get

= 9. 8m + 64k + 64

By taking out the common terms

= 8 (9m + 8k + 8)

= 8r, where r = (9m + 8k + 8) is a natural number

So 3 ^{2(k + 1) + 2} â€“ 8 (k + 1) â€“ 9 is divisible by 8

P (k + 1) is true whenever P (k) is true.

**23. 41^{n}Â â€“ 14^{n}Â is a multiple of 27**

**Solution:**

We can write the given statement as

P (*n*):41* ^{n}*Â â€“ 14

*is a multiple of 27*

^{n}If n = 1 we get

P (1) = 41^{1} â€“ 14^{1} = 27, which is a multiple by 27

Which is true.

Consider P (k) be true for some positive integer k

41* ^{k}*Â â€“ 14

*is a multiple of 27*

^{k}41* ^{k}*Â â€“ 14

*Â = 27*

^{k}*m*, whereÂ

*m*Â âˆˆÂ N

**Â**â€¦â€¦ (1)

Now let us prove that P (k + 1) is true.

Here

41^{k + 1} â€“ 14 ^{k + 1}

We can write it as

= 41^{k}. 41 â€“ 14^{k}. 14

By adding and subtracting 14^{k} we get

= 41 (41^{k} â€“ 14^{k} + 14^{k}) â€“ 14^{k}. 14

On further simplification

= 41 (41^{k} â€“ 14^{k}) + 41. 14^{k} â€“ 14^{k}. 14

From equation (1) we get

= 41. 27m + 14^{k} ( 41 â€“ 14)

By multiplying the terms

= 41. 27m + 27. 14^{k}

By taking out the common terms

= 27 (41m â€“ 14^{k})

= 27r, where r = (41m â€“ 14^{k}) is a natural number

So 41^{k + 1} â€“ 14^{k + 1} is a multiple of 27

P (k + 1) is true whenever P (k) is true.

**24. (2 nÂ +7) < (nÂ + 3)^{2}**

**Solution:**

We can write the given statement as

P(*n*): (2*nÂ *+7) < (*n*Â + 3)2

If n = 1 we get

2.1 + 7 = 9 < (1 + 3)^{2}Â = 16

Which is true.

Consider P (k) be true for some positive integer k

(2*k*Â + 7) < (*k*Â + 3)^{2}Â â€¦ (1)

Now let us prove that P (k + 1) is true.

Here

{2 (k + 1) + 7} = (2k + 7) + 2

We can write it as

= {2 (k + 1) + 7}

From equation (1) we get

(2k + 7) + 2 < (k + 3)^{2} + 2

By expanding the terms

2 (k + 1) + 7 < k^{2} + 6k + 9 + 2

On further calculation

2 (k + 1) + 7 < k^{2} + 6k + 11

Here k^{2} + 6k + 11 < k^{2} + 8k + 16

We can write it as

2 (k + 1) + 7 < (k + 4)^{2}

2 (k + 1) + 7 < {(k + 1) + 3}^{2}

P (k + 1) is true whenever P (k) is true.

### NCERT Solutions for Class 11 Maths Chapter 4- Principle of Mathematical Induction

This chapter has only one exercise which will help students in understanding the concepts related to the Principle of Mathematical Induction clearly. The major topic and subtopics covered in Chapter 4Â Principle of Mathematical Induction of NCERT Solutions for Class 11 include the following.**4.1 Introduction**

Here, students can understand deductive reasoning with suitable examples. This section explains the assumptions that are made on the basis of certain universal facts.Â **4.2 Motivation**

In this section, mathematical induction is explained with a real-life scenario to make the students understand how it basically works.**4.3 The Principle of Mathematical Induction**

This section explains the Principle of Mathematical Induction using inductive step and the inductive hypothesis.

Suppose there is a given statement P(n) involving the natural number n such that

- The statement is true for n = 1, i.e., P(1) is true
- If the statement is true for n = k (where k is some positive integer), then the statement is also true for n = k + 1, i.e., the truth of P(k) implies the truth of P (k + 1).

### Key Features of NCERT Solutions for Class 11 Maths Chapter 4- Principle of Mathematical Induction

Studying the Principle of Mathematical Induction of **Class 11** enables the students to understand the process of the proof by induction, motivating the application of the method by looking at natural numbers as the least inductive subset of real numbers. Students also get to know the principle of mathematical induction and its applications after going through the solutions of NCERT questions. The summary of the concepts discussed and used in the solutions of this chapter are:

- One key basis for mathematical thinking is deductive reasoning. In contrast to deduction, inductive reasoning depends on working with different cases and developing a conjecture by observing incidences until we have observed each and every case. Thus, in simple language we can say the word â€˜inductionâ€™ means the generalization from particular cases or facts
- The principle of mathematical induction is one such tool which can be used to prove a wide variety of mathematical statements. Each such statement is assumed as P(n) associated with positive integer n, for which the correctness of the case n = 1 is examined. Then assuming the truth of P(k) for some positive integer k, the truth of P (k+1) is established.