Question -
Answer -
Let us consider P (n) = 12 + 22 +32 + … + n2 = [n (n+1) (2n+1)]/6
For, n = 1
P (1) = [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 = [k (k+1) (2k+1)]/6
Let’s check for P (n) = k + 1, so
P (k) = 12 + 22 + 32 +– – – – – + k2 + (k + 1)2 = [k + 1 (k+2)(2k+3)] /6
= 12 + 22 + 32 +– – – – – + k2 + (k + 1)2
= [k + 1 (k+2) (2k+3)] /6 + (k + 1)2
= (k +1) [(2k2 + k)/6 + (k + 1)/1]
= (k +1) [2k2 + k + 6k + 6]/6
= (k +1) [2k2 + 7k + 6]/6
= (k +1) [2k2 + 4k + 3k + 6]/6
= (k +1) [2k(k + 2) + 3(k + 2)]/6
= [(k +1) (2k + 3) (k + 2)] / 6
P (n) is true for n = k + 1
Hence, P (n) is true for all n ∈ N.