Show that 1^2 + 2^2 + 3^2 …n^2 = n(n+1)(2n+1)/6

We can prove this by
induction.

For n = 1, n (n+1) (2n+1)/6 = 1*2*3/6 = 1.
Therefore the relation is true.

Now, if we assume 1^2 + 2^2
+ 3^2 …n^2 = n (n+1) (2n+1)/6

1^2 + 2^2 + 3^2 …n^2 + (n +1)
^2 = n (n+1) (2n+1)/6 + (n+1) ^2

= (n+1) [n (2n+1)/6 +
n+1]

= (n +1) (2n^2 + n + 6n +
6)/6

= (n +1) (2n^2 + 7n +
6)/6

= (n +1) (2n^2 + 3n + 4n +
6)/6

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

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

= (n +1) (n+1+ 1) (2(n +1)
+1)/6

Which is the expression n (n+1) (2n+1)/6, with n
replaced by n+1.

So as the relation is true for n = 1, and
if it is assumed true for any n, we can show that it is also true for n+1; the relation
is true for all values of n.