NCERT Solutions for Class 11 Maths Chapter 4 - Principle of Mathematical Induction Exercise 4.1

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

NCERT Solutions are provided to help the students understand the steps to solve mathematical problems that are provided in the textbook. Exercise 4.1 of NCERT Solutions for Class 11 Maths Chapter 4 – Principle of Mathematical Induction is the only exercise in this chapter. It includes questions from all the topics covered in this chapter.

1. 1. Motivation
   2. The Principle of Mathematical Induction

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

1. • 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).

The subject specialists stick to the syllabus while preparing the solutions. The problem-solving method provided in the examples is followed while curating the NCERT Solutions for Class 11.

NCERT Solutions for Class 11 Maths Chapter 4 – Principle of Mathematical Induction Exercise 4.1

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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) to 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) to 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) to 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) to 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) to 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) to 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) to 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.