# Class 11 RD Sharma Solutions – Chapter 12 Mathematical Induction – Exercise 12.1

• Last Updated : 11 Dec, 2020

### Question 1: If P (n) is the statement “n (n + 1) is even”, then what is P (3)?

Solution:

Given: P (n) = n (n + 1) is even.

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Substituting n with 3 we get,

P (3) = 3 (3 + 1)

P (3) = 3 (4)

P (3) = 12

Since P (3) = 12, and 12 is even

Therefore, P (3) is also even.

### Question 2: If P (n) is the statement “n3 + n is divisible by 3”, prove that P (3) is true but P (4) is not true.

Solution:

Given: P (n) = n3 + n is divisible by 3

Firstly, substituting n with 3 we get,

P (3) = 33 + 3

P (3) = 27 + 3

P (3) = 30

Since P (3) = 30, and 30 is divisible by 3

Therefore, P (3) is true.

Now, substituting n with 4 we get,

P (4) = 43 + 4

P (4) = 64 + 4

P (4) = 68

Since P (4) = 68, and 68 it is not divisible by 3

### Question 3: If P (n) is the statement “2n ≥ 3n”, and if P (r) is true, prove that P (r + 1) is true.

Solution:

Given: P (n) = 2n ≥ 3n and p(r) is true.

also given that P (r) is true

When we substitute n with r we get

P (r) = 2r ≥ 3r

Now, multiply both sides by 2 we get,

2 × 2r ≥ 3r × 2

2r + 1 ≥ 6r

We can write 6r as 3r + 3r

2r + 1 ≥ 3r + 3r

Since 3r ≥ 3, therefore 3r + 3r ≥ 3 + 3r

Substituting  3r + 3r with 3 + 3r we get,

2r + 1 ≥ 3 + 3r

2r + 1 ≥ 3(r + 1)      [Taking 3 as common]

Since, 2r+1 ≥ 3(r + 1) is equal to P (r + 1)

Therefore, P (r + 1) is true.

### Question 4: If P (n) is the statement “n2 + n is even”, and if P (r) is true, then P (r + 1) is true

Solution:

Given: P (n) = n2 + n is even

Also given that P (r) is true,

Therefore, P (r) = r2 + r is even

Let us consider r2 + r = 2x … (i)

Substituting r with r + 1

Now, (r + 1)2 + (r + 1)

r2 + 1 + 2r + r + 1    [ formula = (a + b)2 = a2 + 2ab + b2]

(r2 + r) + 2r + 2

2x + 2r + 2 [from equation (i) we get 2x = r2 + r]

2(x + r + 1)

Since, (r + 1)2 + (r + 1) is Even which is equal to P (r + 1).

Therefore, P (r + 1) is true.

### Question 5: Given an example of a statement P (n) such that it is true for all n ϵ N.

Solution:

Let us consider P (n) as

P (n) = 1 + 2 + 3 + – – – – – + n = n(n+1)/2

Since P (n) is true for all natural numbers.

Therefore, P (n) is true for all n ∈ N.

### Question 6: If P (n) is the statement “n2 – n + 41 is prime”, prove that P (1), P (2), and P (3) are true. Prove also that P (41) is not true.

Solution:

Given: P(n) = n2 – n + 41 is prime.

Substituting n with 1, we get

P (1) = 1 – 1 + 41

P (1) = 41

Since P (1) = 41, and 41 is prime.

Therefore, P (1) is true.

Now substituting n with 2, we get

P(2) = 22 – 2 + 41

P(2) = 4 – 2 + 41

P(2) = 43

Since P (2) = 43, and 43 is prime.

Therefore, P (2) is true.

Now substituting n with 3, we get

P (3) = 32 – 3 + 41

P (3) = 9 – 3 + 41

P (3) = 47

Since P (3) = 47, and 47 is prime.

Therefore, P (3) is true.

Now substituting n with 41, we get

P (41) = (41)2 – 41 + 41

P (41) = 1681

Since P (41) = 1681, and 1681 is not prime.

Therefore, P (41) is not true.

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