Multiplying $P(x) = (x-1)(x-2) \dots (x-50)$ and $Q(x)=(x+1)(x+2) \cdots(x+50)$ 
Let $P(x) = (x-1)(x-2) \dots (x-50)$ and $Q(x)=(x+1)(x+2) \cdots(x+50).$
If $P(x)Q(x) = a_{100}x^{100} + a_{99}x^{99} + \dots + a_{1}x^{1} + a_0$, compute $a_{100} - a_{99} - a_{98} - a_{97}.$

I've been quite stuck with this one. If I multiply and group the polynomials with the similar terms e.g $(x-1)$ and $(x+1) ...$ I would get
$P(x)Q(x)= (x^2-1)(x^2-2^2) \dots(x^2-50^2)$ right? From here on if I would keep multiplying the terms wouldn't I end up with $50$ times the term $x^2$ as the initial term, hence $a_{100} = 1$ since it wouldn't have any coefficient? For the other terms I don't have a clue how to find them so any hints would be appreciated
 A: Clearly $a_{99}=a_{97}=0$, since only even degree terms appear. As you've noticed, $a_{100}=1$; so really we just need to find $a_{98}$. But this is just negative $1$ times the sum of squares up to $50$, i.e. $-\sum_{k=1}^{50} k^2 =- \frac{50\cdot 51\cdot 101}{6}=-42925$. So the answer is $42926$.
Edit: it seems I was a bit sparse with my explanation. Claim: the coefficient of $x^{2n-2}$ in $\prod_{i=1}^{n}(x^2-i^2)$ is $\sum_{i=1}^{n}i^2 = \frac{n(n+1)(2n+1)}{6}$. The claim is immediate when $n=1$ or $n=2$:
$$
(n=1)\qquad \prod_{i=1}^{1}(x^2-i^2)= x^2-1;\, 1 = \frac{1\cdot 2 \cdot 3}{6}
$$
$$
(n=2)\qquad \prod_{i=1}^{2}(x^2-i^2)= x^4-5x^2+1;\, 5 = \frac{2\cdot 3 \cdot 5}{6}
$$Suppose the result is true for some $k$. Then
$$
\prod_{i=1}^{k+1 }(x^2-i^2) = \left(\prod_{i=1}^{k }(x^2-i^2)\right)\cdot(x^2-(k+1)^2)
$$
$$
 = \left(x^{2k}-\sum_{i=1}^{k}i^2 \cdot x^{2k-2} +\cdots\right)\cdot(x^2-(k+1)^2)
$$
$$
 = \left(x^{2k+2}-((k+1)^2+\sum_{i=1}^{k}i^2) \cdot x^{2k} +\cdots\right),
$$as was to be shown.
