Let P (n): n (n + 1) (n + 5) is a multiple of 3

Let us check for n = 1,

P (1): 1 (1 + 1) (1 + 5)

: 2 × 6

: 12

P (n) is true for n = 1. Where, P (n) is a multiple of 3

Now, let us check for P (n) is true for n = k, and have to prove that P (k + 1) is true.

P (k): k (k + 1) (k + 5) is a multiple of 3

: k(k + 1) (k + 5) = 3λ … (i)

We have to prove,

(k + 1)[(k + 1) + 1][(k + 1) + 5] is a multiple of 3

(k + 1)[(k + 1) + 1][(k + 1) + 5] = 3μ

So,

= (k + 1) [(k + 1) + 1] [(k + 1) + 5]

= (k + 1) (k + 2) [(k + 1) + 5]

= [k (k + 1) + 2(k + 1)] [(k + 5) + 1]

= k (k + 1) (k + 5) + k(k + 1) + 2(k + 1) (k + 5) + 2(k + 1)

= 3λ + k^{2} + k + 2(k^{2} + 6k + 5) + 2k + 2

= 3λ + k^{2} + k + 2k^{2} + 12k + 10 + 2k + 2

= 3λ + 3k^{2} + 15k + 12

= 3(λ + k^{2} + 5k + 4)

= 3μ

P (n) is true for n = k + 1

Hence, P (n) is true for all n ∈ N.

Answered by Aaryan | 1 year agoGiven an example of a statement P (n) such that it is true for all n ϵ N.

a + (a + d) + (a + 2d) + … + (a + (n-1)d) = \( \dfrac{n}{2}\) [2a + (n-1)d]

7^{2n} + 2^{3n – 3}. 3n – 1 is divisible by 25 for all n ϵ N

(ab)^{ n} = a^{n} b^{n} for all n ϵ N

3^{2n + 2} – 8n – 9 is divisible by 8 for all n ϵ N