Find the sum to n terms of the series 12 + 32 + 52 + ... to n terms.

We know that, 12+22+......+n2=n(n+1)(2n+1)6

Here, 

t1=12=(21-1)2

t2=32=(22-1)2

Hence in this way the n th term is, tn=(2n-1)2

So, 

12+32+52+....+(2n-1)2

={12+22+32+42+52+......+(2n-2)2+(2n-1)2+(2n)2}-{22+42+...+(2n-2)2+(2n)2}

=2n(2n+1)(22n+1)6-22{12+22+...+(n-1)2+n2}

=n(2n+1)(4n+1)3-4n(n+1)(2n+1)6

=n(2n+1)(4n+1)3-2n(n+1)(2n+1)3

=n(2n+1)3[4n+1-2(n+1)]

=n(2n+1)3[4n+1-2n-2]

=n(2n+1)(2n-1)3

Hence the sum of n terms of the series 12+32+52+.... is n(2n+1)(2n-1)3.