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

**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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free classeswhich will definitely help them in making a wise career choice in the future.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 “n**^{3} + n is divisible by 3”, prove that P (3) is true but P (4) is not true.

^{3}+ n is divisible by 3”, prove that P (3) is true but P (4) is not true.

**Solution:**

Given: P (n) = n

^{3}+ n is divisible by 3Firstly, substituting n with 3 we get,

P (3) = 3

^{3}+ 3P (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) = 4

^{3}+ 4P (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 “2**^{n} ≥ 3n”, and if P (r) is true, prove that P (r + 1) is true.

^{n}≥ 3n”, and if P (r) is true, prove that P (r + 1) is true.

**Solution:**

Given: P (n) = 2

^{n}≥ 3n and p(r) is true.also given that P (r) is true

When we substitute n with r we get

P (r) = 2

^{r }≥ 3rNow, multiply both sides by 2 we get,

2 × 2

^{r }≥ 3r × 22

^{r + 1 }≥ 6rWe can write 6r as 3r + 3r

2

^{r + 1}≥ 3r + 3rSince 3

^{r}≥ 3, therefore 3r + 3r ≥ 3 + 3rSubstituting 3r + 3r with 3 + 3r we get,

2

^{r + 1}≥ 3 + 3r2

^{r + 1}≥ 3(r + 1) [Taking 3 as common]Since, 2

^{r+1}≥ 3(r + 1) is equal to P (r + 1)Therefore, P (r + 1) is true.

**Question 4: If P (n) is the statement “n**^{2} + n is even”, and if P (r) is true, then P (r + 1) is true

^{2}+ n is even”, and if P (r) is true, then P (r + 1) is true

**Solution:**

Given: P (n) = n

^{2}+ n is evenAlso given that P (r) is true,

Therefore, P (r) = r

^{2}+ r is evenLet us consider r

^{2}+ r = 2x … (i)Substituting r with r + 1

Now, (r + 1)

^{2}+ (r + 1)r

^{2}+ 1 + 2r + r + 1 [ formula = (a + b)^{2}= a^{2}+ 2ab + b^{2}](r

^{2}+ r) + 2r + 22x + 2r + 2 [from equation (i) we get 2x = r

^{2}+ r]2(x + r + 1)

2μ

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 “n**^{2} – n + 41 is prime”, prove that P (1), P (2),** and P (3) are true. Prove also that P (41) is not true.**

^{2}– n + 41 is prime”, prove that P (1), P (2)

**Solution:**

Given: P(n) = n

^{2}– 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) = 2

^{2}– 2 + 41P(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) = 3

^{2}– 3 + 41P (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 + 41P (41) = 1681

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

Therefore, P (41) is not true.