Combinational Sum!!

Let , $P(x)=(x-1)(x+1)(x+2)(x-2)\dots \dots (x-50)(x+50)$

$\Rightarrow P(x)=(x^2-1)(x^2-2^2) \dots \dots (x^2-50^2)$

All roots of $P(x)$ are elements of set $S$ .

Now by veita, we can easily realize that $\sum \limits_{i \in A}{i}$ is coefficient of $x^{97}$ .

But $P(x)$ (after expanding) has no odd degree term. So the answer is $\fbox0$

Firstly, we can ignore all subsets that contain 0, because they don't affect the sum. Now, note that for any subset, we can negate one of the terms to get a different subset, which cancels with the sum. Therefore, the final sum is 0.