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.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)

= 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

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.