Answer
$\displaystyle \sum_{n=1}^{\infty}(n^{2}-2)$
Work Step by Step
We first start with the perfect squares:
1, 4, 9, 16, 25, ... $\qquad $(squares of 1,2,3,...)
Starting with n=1, the terms in the sum are
$a_{n}=n^{2}-2$,
and there is no upper limit on n.
Sum = $\displaystyle \sum_{n=1}^{\infty}(n^{2}-2)$
