Coefficient Problem (polynomial expansion) 
Let $C$ be the coefficient of $x^2$ in the expansion of the product $(1 - x)(1 + 2x)(1 - 3x)\cdots(1 + 14x)(1 - 15x).$ Find $|C|.$

Just to begin,
$(1-x)(1+2x) = -2x^2 + x + 1$
$(1-x)(1+2x)(1-3x) = 6x^3 - 5x^2 - 2x + 1$ 
But expanding on like this take too long.
In the end the terms will be like:
$(1 - 15x)(a_0 + a_1x^1 + a_2x^2 + ....)$
Then just looking at the $x^2$, it will be:
$(a_2x^2 - 15a_1x^2) = x^2(a_2 - 15a_1)$
But it it still a very hard problem, any hints?

We get the coefficient of $x^2$ by adding the product of pairs of coefficients of $x$ in the binomials, since the other coefficients would be $1$.
All the coefficients that come from the $1-x$ term multiplied by other $x$ terms add up to
$$-1\cdot 2 + -1\cdot -3 + \ldots + -1\cdot -15$$
$$=-1(2-3+\ldots -15)$$
$$=-1[(-1+2-3+\ldots -15)+1]$$
$$=-1(T+1)$$
$$=-1T-1^2$$
where $T=-1+2-3+\ldots -15$. The coefficients that come from the $1+2x$ term multiplied by other $x$ terms add up to
$$2\cdot -1 + 2\cdot -3 + \ldots + 2\cdot -15$$
$$=2(-1-3+\ldots -15)$$
$$=2[(-1+2-3+\ldots -15)-2]$$
$$=2(T-2)$$
$$=2T-2^2$$
I think you get the idea. The coefficients coming from the $1-15x$ term multiplied by other $x$ terms add up to
$$-15T-15^2$$
Adding all those, to get the twice the sum of all products of pairs of $x$ coefficients (twice since we got each pair twice), we get
$$2C=(-1+2-3+\ldots -15)T-(1^2+2^2+3^2+\ldots +15^2)$$
$$=T^2-(1^2+2^2+3^2+\ldots +15^2)$$
So calculate $T$ and the sum of the first $15$ square numbers, substitute, and you are done. $T$ is easy:
$$T=-1+(2-3)+(4-5)+\ldots +(14-15)$$
$$=-1+-1+-1+\ldots +-1 \quad\text{($8$ times)}$$
$$=-8$$
The sum of the squares is
$$\frac{15\cdot (15+1)(2\cdot 15+1)}6=1240$$
So $C=\frac{(-8)^2-1240}2=-588$.
(Sorry for the late completion of this answer: all internet access in this area was cut off for a while.)
