Question

What is the sum of series if nth term is (2n+1)^2

Answer 1

http://www.wolframalpha.com/input/?i=summation++%282n%2B1%29^2

Answer 2

It is \[\Sigma n\] where n is nth term ie,(\[\Sigma (2n+1)^2=\Sigma (4n^2+4n+1)=4\Sigma (n^2)+4\Sigma (n)+\Sigma (1)\]
We have \[\Sigma (n^2)= n(n+1)(2n+1)/6\]
\[\Sigma (n)=n(n+1)/2\]
\[ \Sigma(1)=n\]
Substituting these in above expression and simplify to get the answer.

Answer 3

Answer is \[(4n^3+12n^2+11n)/6\]