Let us check for n = 1,

P (1): 1.3 = \( \dfrac{1}{6}\) × 1 × 2 × 9

3 = 3

P (n) is true for n = 1.

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

P (k): 1.3 + 2.4 + 3.5 + … + k. (k+2) = \( \dfrac{1}{6}\) k (k+1) (2k+7) … (i)

So,

1.3 + 2.4 + 3.5 + … + k. (k+2) + (k+1) (k+3)

Now, substituting the value of P (k) we get,

= \( \dfrac{1}{6}\)k (k+1) (2k+7) + (k+1) (k+3) by using equation (i)

=\( (k+1) \dfrac{k(2k+7)}{6} + \dfrac{(k+3)}{1}\)

= \(\dfrac{ (k+1) (2k+9) (k+2) }{ 6}\)

= \( \dfrac{1}{6}\)(k+1) (k+2) (2k+9)

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

n (n + 1) (n + 5) is a multiple of 3 for all n ϵ N.

(ab)^{ n} = a^{n} b^{n} for all n ϵ N