NCERT Solutions For Class 11 Maths Chapter 4 Principle of Mathematical Induction

*According to the latest update on the CBSE Syllabus 2023-24, this chapter has been removed.

NCERT Solutions For Class 11 Maths Chapter 4 Principle of Mathematical Induction is given in an understandable way by the faculty at BYJU’S. Students learn about the Principle of Mathematical Induction and its application in detail through this chapter. By practising all the problems present in the NCERT Solutions, students can easily score maximum marks in the examinations.
The 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 BYJU’S cover all these concepts and help in scoring full marks in 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 for the students to crack the exam with good marks.

NCERT Solutions For Class 11 Maths Chapter 4 Principle of Mathematical Induction

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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 ∈ 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.

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

5.

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.

6.

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.

7.

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.

8. 1.2 + 2.2² + 3.2² + … + n.2ⁿ = (n – 1) 2ⁿ⁺¹ + 2

Solution:

We can write the given statement as

P (n): 1.2 + 2.2² + 3.2² + … + n.2ⁿ = (n – 1) 2ⁿ⁺¹ + 2

If n = 1 we get

P (1): 1.2 = 2 = (1 – 1) 2¹⁺¹ + 2 = 0 + 2 = 2

Which is true.

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

1.2 + 2.2² + 3.2² + … + k.2^k = (k – 1) 2^{k + 1} + 2 … (i)

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

Here

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.

9.

Solution:

We can write the given statement as

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.

10.

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.

11.

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.

12.

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.

13.

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.

14.

Solution:

By further simplification

= (k + 1) + 1

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.

15.

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.

16.

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.

17.

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.

18.

Solution:

We can write the given statement as

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.

19. n (n + 1) (n + 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 ∈ N …… (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 × 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.

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

20. 10^{2n – 1} + 1 is divisible by 11

Solution:

We can write the given statement as

P (n): 10^{2n – 1} + 1 is divisible by 11

If n = 1 we get

P (1) = 10^{2.1 – 1} + 1 = 11, which is divisible by 11

Which is true.

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

10^{2k – 1} + 1 is divisible by 11

10^{2k – 1} + 1 = 11m, where m ∈ N …… (1)

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

Here

10 ^{2 (k + 1) – 1} + 1

We can write it as

= 10 ^{2k + 2 – 1} + 1

= 10 ^{2k + 1} + 1

By addition and subtraction of 1

= 10 ² (10^2k-1 + 1 – 1) + 1

We get

= 10 ² (10^2k-1 + 1) – 10² + 1

Using equation 1 we get

= 10². 11m – 100 + 1

= 100 × 11m – 99

Taking out the common terms

= 11 (100m – 9)

= 11 r, where r = (100m – 9) is some natural number

10 ^{2(k + 1) – 1} + 1 is divisible by 11

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.

21. x²ⁿ – y²ⁿ is divisible by x + y

Solution:

We can write the given statement as

P (n): x²ⁿ – y²ⁿ is divisible by x + y

If n = 1 we get

P (1) = x^2 × 1 – y^2 × 1 = x² – y² = (x + y) (x – y), which is divisible by (x + y)

Which is true.

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

x^2k – y^2k is divisible by x + y

x^2k – y^2k = m (x + y), where m ∈ N …… (1)

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

Here

x ^{2(k + 1)} – y ^{2(k + 1)}

We can write it as

= x ^2k . x² – y^2k . y²

By adding and subtracting y^2k we get

= x² (x^2k – y^2k + y^2k) – y^2k. y²

From equation (1) we get

= x² {m (x + y) + y^2k} – y^2k. y²

By multiplying the terms

= m (x + y) x² + y^2k. x² – y^2k. y²

Taking out the common terms

= m (x + y) x² + y^2k (x² – y²)

Expanding using formula

= m (x + y) x² + y^2k (x + y) (x – y)

So we get

= (x + y) {mx² + y^2k (x – y)}, which is a factor of (x + y)

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.

22. 3^{2n + 2} – 8n – 9 is divisible by 8

Solution:

We can write the given statement as

P (n): 3^{2n + 2} – 8n – 9 is divisible by 8

If n = 1 we get

P (1) = 3^{2 × 1 + 2} – 8 × 1 – 9 = 64, which is divisible by 8

Which is true.

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

3^{2k + 2} – 8k – 9 is divisible by 8

3^{2k + 2} – 8k – 9 = 8m, where m ∈ N …… (1)

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

Here

3 ^{2(k + 1) + 2} – 8 (k + 1) – 9

We can write it as

= 3 ^{2k + 2} . 3² – 8k – 8 – 9

By adding and subtracting 8k and 9 we get

= 3² (3^{2k + 2} – 8k – 9 + 8k + 9) – 8k – 17

On further simplification

= 3² (3^{2k + 2} – 8k – 9) + 3² (8k + 9) – 8k – 17

From equation (1) we get

= 9. 8m + 9 (8k + 9) – 8k – 17

By multiplying the terms

= 9. 8m + 72k + 81 – 8k – 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} – 8 (k + 1) – 9 is divisible by 8

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.

23. 41ⁿ – 14ⁿ is a multiple of 27

Solution:

We can write the given statement as

P (n):41ⁿ – 14ⁿis a multiple of 27

If n = 1 we get

P (1) = 41¹ – 14¹ = 27, which is a multiple by 27

Which is true.

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

41^k – 14^kis a multiple of 27

41^k – 14^k = 27m, where m ∈ N …… (1)

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

Here

41^{k + 1} – 14 ^{k + 1}

We can write it as

= 41^k. 41 – 14^k. 14

By adding and subtracting 14^k we get

= 41 (41^k – 14^k + 14^k) – 14^k. 14

On further simplification

= 41 (41^k – 14^k) + 41. 14^k – 14^k. 14

From equation (1) we get

= 41. 27m + 14^k ( 41 – 14)

By multiplying the terms

= 41. 27m + 27. 14^k

By taking out the common terms

= 27 (41m – 14^k)

= 27r, where r = (41m – 14^k) is a natural number

So 41^{k + 1} – 14^{k + 1} is a multiple of 27

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.

24. (2n +7) < (n + 3)²

Solution:

We can write the given statement as

P(n): (2n +7) < (n + 3)2

If n = 1 we get

2.1 + 7 = 9 < (1 + 3)² = 16

Which is true.

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

(2k + 7) < (k + 3)² … (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 < k² + 6k + 9 + 2

On further calculation

2 (k + 1) + 7 < k² + 6k + 11

Here k² + 6k + 11 < k² + 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.

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

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

This chapter has only one exercise which will help students to understand the concepts related to the Principle of Mathematical Induction clearly. The major topic and subtopics covered in Chapter 4  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.
4.2 Motivation
In this section, mathematical induction is explained with a real-life scenario to help 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:

1. 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 ‘induction’ means the generalization from particular cases or facts
2. 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.