Answer to Question #9058 in Discrete Mathematics for ashok

Question

Prove by principle of Mathematical induction that sum of squares of first n natural
number is {n(n+1)(2n+1)}/6

Answer 1

Let n=1
Then
{n(n+1)(2n+1)}/6 = 1*2*3/6 = 1
and

1^2=1
as well.

Suppose that we have proved that
1^2 + 2^2 + ...
+ k^2 = {k(k+1)(2k+1)}/6
for all k<n+1
Let us prove this identity for
k=n+1, that is
1^2 + 2^2 + ... + n^2 + (n+1)^2 = {(n+1)(n+2)(2(n+1)+1)}/6

= {(n+1)(n+2)(2n+3)}/6

Consider
the sum
S = 1^2 + 2^2 + ... + n^2 + (n+1)^2 ** by inductive
step
= {n(n+1)(2n+1)}/6 + (n+1)^2
= [ n(2n+1) +
6(n+1) ] * (n+1)/6
= [ 2n^2 + n + 6n + 6 ] * (n+1)/6
= [
2n^2 + 7n + 6 ] * (n+1)/6
= [ (n+2)(2n+3) ] * (n+1)/6
=
{(n+1)(n+2)(2n+3)}/6