12 + 22 + 32 + … + n2 = [n (n+1) (2n+1)]/6

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): 1² + 2² + 3² + … + k² =\(\dfrac{ [k (k+1) (2k+1)]}{6}\)

Let’s check for P (n) = k + 1, so

P (k) = 1² + 2² + 3² + – – – – – + k² + (k + 1)² = \(\dfrac{ [k + 1 (k+2) (2k+3)] }{6}\)

= 1² + 2² + 3² + – – – – – + k² + (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.