Continuous practice using ML Aggarwal Class 11 ISC Solutions Chapter 4 Principle of Mathematical Induction Ex 4.1 can lead to a stronger grasp of mathematical concepts.

## ML Aggarwal Class 11 Maths Solutions Section A Chapter 4 Principle of Mathematical Induction Ex 4.1

Question 1.

If P (n) is the statement “n (n + 1) (n + 2) is divisible by 6”, then what is P (3)?

Solution:

P (n) is the statement “n (n + 1) (n + 2) is divisible by 6″

∴ P(3) is 3 (3 + 1) (3 + 2) is divisible by 6

i.e. 60 is divisible by 6. Which is true.

Question 2.

If P (n) is the statement “10n + 3 is prime”, then show that P (1) and P (2) are true but P (3) is not true.

Solution:

Given P (n) is the statement “10n + 3 is prime”

∴ p (1) i.e. 10 × 1 + 3 is prime i.e. 13 is prime, which is true.

So P (1) is true.

p (2) i.e. 10 × 2 + 3 i.e. 23 is prime, which is true.

P (2) is true.

p (3) i.e. 10 × 3 + 3 is prime i.e. 33 is prime

which is not true since 3 and 11 are factors of 33.

Thus, P (3) is not true.

Question 3.

If P (n) is the statement “n (n + 1) (n + 2) is an integral multiple of 12”, prove that P (3) and P (4) are true but P (5) is not true.

Solution:

Given P (n) is the statement “n (n + 1 )(n+2) is an integral multiple of 12”.

Now P(3) i.e. 3(3 + 1) (3 +2) is an integral multiple of 12.

∴ 60 is an integral multiple of 12, which is clearly true.

Thus, P (3) is true.

Now P (4) i.e. 4 (4 + 1) (4 + 2) is an integral multiple of 12.

∴ 4.5.6 – 120 is an integral multiple of 12

which is clearly true.

Here P (5) i.e. 5 (5 + 1) (5 + 2) is an integral multiple of 12.

∴ 5 × 6 × 7 = 210 is an integral multiple of 12 which is not true.

Thus, P (5) is not true.

Question 4.

If P (n) is the statement “n^{2} – n + 41 is prime”, show that P (1), P (2), and P (3) are true but P (41) is not true.

Solution:

Given P (n) is the statement ‘n^{2} – n + 41 is prime”.

∴ P(1) is 1^{2} – 1 + 41 is prime

⇒ 41 is prime, which ¡s clearly true.

P (2) is 2^{2} – 2 + 41 is prime

∴ 43 is prime, which is clearly true.

P (3) is 3^{2} – 3 + 41 is prime.

∴ 47 is prime, which is clearly true.

and P (41) is 41^{2} – 41 + 41 is prime i.e.

(41)^{2} is prime

which is not true since 41 be a factor of (41)^{2}.

Thus, P (41) is not title.

Question 5.

Let P (n) is the statement “n^{2} + n is an even integer”. Show that if P (k) is true then P (k + 1) is also true.

Solution:

Given P (n) is the statement “n^{2} + n is an even integer”.

Let P (k) is true k^{2} + k is an even integer.

⇒ k^{2} + k = 2λ where λ ∈ I

Now P( k + 1) = (k + 1)^{2} + k + 1

⇒ k^{2} + 2k + 1 + k + 1

= (k^{2} + k) + 2k + 2

= 2λ + 2k + 2

= 2 (λ + k + 1)

= even integer

Thus P (k + 1) also true.

Question 6.

If P (n) denote the statement “3^{2n} – 1 is a multiple of 8”. Show that

(i) P (1), P (2) are true

(ii) if P (m) is true then P (m + 1) is also true.

Solution:

Given P (n) denote the statement “3^{2n} – 1 is a multiple of 8”.

(i) ∴ P (1) is 3^{2 × 1} – 1 i.e. 8 is a multiple of 8, which is true.

Thus, P (1) is true.

and P (2) is 3^{2 × 1} – 1 = 81 – 1 = 80 is a multiple of 8

which is clearly true.

(ii) Let P (m) is true

⇒ 3^{2m} – 1 is a multiple of 8

⇒ 3^{2m} – 1 = 8λ for some integer λ

⇒ 3^{2m} = (8λ + 1) …………………(1)

Now, P (m + 1) = 3^{2 (m + 1)} – 1

= 3^{2m} . 3^{2} – 1

= 9 (8λ + 1) – 1 [using (1)]

= 72λ + 8

= 8 (9λ + 1)

which is a multiple of 8.

Thus, P (m + 1) is true.