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

NCERT Solutions are provided to help the students in understanding 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. Motivation
2. The Principle of Mathematical Induction:

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).

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.

## Download PDF of NCERT Solutions for Class 11 Maths Chapter 4- Principle of Mathematical Induction Exercise 4.1                           #### Access Solutions for Class 11 Maths Chapter 4.1 Exercise

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.2k = (k – 1) 2k + 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. 102n – 1 + 1 is divisible by 11

Solution:

We can write the given statement as

P (n): 102n – 1 + 1 is divisible by 11

If n = 1 we get

P (1) = 102.1 – 1 + 1 = 11, which is divisible by 11

Which is true.

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

102k – 1 + 1 is divisible by 11

102k – 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 (102k-1 + 1 – 1) + 1

We get

= 10 2 (102k-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) = x2 × 1 – y2 × 1 = x2 – y2 = (y) (x – y), which is divisible by (x + y)

Which is true.

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

x2k – y2k is divisible by x y

x2k – y2k = m (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 – y2k . y2

By adding and subtracting y2k we get

= x2 (x2k – y2k + y2k) – y2k. y2

From equation (1) we get

= x2 {m (x + y) + y2k} – y2k. y2

By multiplying the terms

= m (x + y) x2 + y2k. x2 – y2k. y2

Taking out the common terms

= m (x + y) x2 + y2k (x2 – y2)

Expanding using formula

= m (x + y) x2 + y2k (x + y) (x – y)

So we get

= (x + y) {mx2 + y2k (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. 32n + 2 – 8n – 9 is divisible by 8

Solution:

We can write the given statement as

P (n): 32n + 2 – 8n – 9 is divisible by 8

If n = 1 we get

P (1) = 32 × 1 + 2 – 8 × 1 – 9 = 64, which is divisible by 8

Which is true.

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

32k + 2 – 8k – 9 is divisible by 8

32k + 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 (32k + 2 – 8k – 9 + 8k + 9) – 8k – 17

On further simplification

= 32 (32k + 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

41k – 14kis a multiple of 27

41k – 14k = 27m, where m ∈ N …… (1)

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

Here

41k + 1 – 14 k + 1

We can write it as

= 41k. 41 – 14k. 14

By adding and subtracting 14k we get

= 41 (41k – 14k + 14k) – 14k. 14

On further simplification

= 41 (41k – 14k) + 41. 14k – 14k. 14

From equation (1) we get

= 41. 27m + 14k ( 41 – 14)

By multiplying the terms

= 41. 27m + 27. 14k

By taking out the common terms

= 27 (41m – 14k)

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

So 41k + 1 – 14k + 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. (2+7) < (n + 3)2

Solution:

We can write the given statement as

P(n): (2+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.