For, n = 1

P (1) = \( \dfrac{[1 (1+1) (2+1)]}{6}\)

1 = 1

P (n) is true for n = 1

Let P (n) is true for n = k, so

P (k): 1^{2} + 2^{2} + 3^{2} + … + k^{2} =\(\dfrac{ [k (k+1) (2k+1)]}{6}\)

Let’s check for P (n) = k + 1, so

P (k) = 1^{2} + 2^{2} + 3^{2} + – – – – – + k^{2} + (k + 1)^{2} = \(\dfrac{ [k + 1 (k+2) (2k+3)] }{6}\)

= 1^{2} + 2^{2} + 3^{2} + – – – – – + k^{2} + (k + 1)^{2}

= \(\dfrac{ (k +1) [2k^2 + 7k + 6]}{6}\)

= \(\dfrac{ (k +1) [2k^2 + 4k + 3k + 6]}{6}\)

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

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