Prove using induction that sum of i2 from 1 to n nn12n16Sol

Prove using induction that sum of (i^2) from 1 to n = [n(n+1)(2n+1)]/6

Solution

Put n = 1: 1^2 = 1(1+1)(2*1+1)/6 = 1(2)(3)/6 = 6/6 = 1 = 1^2, so this is true.

Then show that if the identity is true for n, it is true for n + 1.

then Put n = n + 1 :

n(n+1)(2n+1)/6 + (n+1)^2
n(n+1)(2n+1)/6 + n^2 + 2n + 1
2n^3/6 + 3n^2/6 + n/6 + n^2 + 2n + 1
((2n^3 + 3n^2 + n + 6n^2 + 12n + 6)/6
(2n^3 + 9n^2 + 13n + 6)/6
(2n^3 + 3n^2 + 6n^2 + 9n + 4n + 6)/6
((n^2 + 3n + 2)(2n+3))/6
((n+1)(n+2)(2n+3))/6
((n+1)((n+1)+1)(2(n+1)+1))/6

hence proved