summation

not a very tough problem, but no one seems to have replied yet!

TO FIND:- 1^2+3^2+5^2+…….to n terms. --------(S1)
nth term= (2n-1)^2

Now…let S=1^2+2^2+3^2+……..+(2n)^2
= 2n(2n+1)(4n+1)/6
= n(2n+1)(4n+1)/3

let S2= 2^2+4^2+6^2+………..+(2n)^2
= 2^2(1^2+2^2+3^2+……..+n^2)
=4n(n+1)(2n+1)/6
=2n(n+1)(2n+1)/3

so, S=S1-S2
=n(4n^2-1)/3 ------answer.