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

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 “n² – 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 ‘n² – n + 41 is prime”.
∴ P(1) is 1² – 1 + 41 is prime
⇒ 41 is prime, which ¡s clearly true.
P (2) is 2² – 2 + 41 is prime
∴ 43 is prime, which is clearly true.
P (3) is 3² – 3 + 41 is prime.
∴ 47 is prime, which is clearly true.
and P (41) is 41² – 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 “n² + 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 “n² + n is an even integer”.
Let P (k) is true k² + k is an even integer.
⇒ k² + k = 2λ where λ ∈ I
Now P( k + 1) = (k + 1)² + k + 1
⇒ k² + 2k + 1 + k + 1
= (k² + 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 “3²ⁿ – 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 “3²ⁿ – 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 . 3² – 1
= 9 (8λ + 1) – 1 [using (1)]
= 72λ + 8
= 8 (9λ + 1)
which is a multiple of 8.
Thus, P (m + 1) is true.