# NCERT Solutions For Class 11 Math Chapter – 4 Exercise – 4.1

NCERT Solutions For Class 11  Math Chapter -4 Exercise – 4.1

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 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

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

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)

= 8 where = (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.