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?
 A: $$\begin{eqnarray*}[x^2]\prod_{i=1}^{15}\left(1+i(-1)^i x\right)=\sum_{1\leq j<k\leq 15}(-1)^{j+k}jk&=&\frac{1}{2}\left(\left(\sum_{j=1}^{15}(-1)^j j\right)^2-\sum_{j=1}^{15}j^2\right)\\&=&\frac{1}{2}\left((-8)^2-1240\right)\\&=&\color{red}{-588}.\end{eqnarray*} $$
A: Somewhat more generally, let $$\prod_{j=1}^n (1 + (-1)^j j x) = 1 + c(n) x + d(n) x^2 + \ldots$$
so you want $d(15)$.
We have $c(0) = d(0) = 0$ with
$$ (1 + c(n-1) x + d(n-1) x^2)(1 + (-1)^n n x) = 1 + c(n) x + d(n) x^2 + \ldots $$
so that 
$$ \eqalign{c(n) &= c(n-1) + (-1)^n n \cr
            d(n) &= d(n-1) + (-1)^n n c(n-1) \cr} $$
Can you solve these recursions?
A: 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.)
A: Another approach: consider the product
$$(1-x)(1+2x)\cdots(1+(-1)^nnx)=1+a_nx+b_nx^2+\cdots\ .$$
It is easy to see that the coefficient of $x$ in this expression is
$$\eqalign{a_n
  &=-1+2-3+4-\cdots+(-1)^nn\cr
  &=\cases{(-1+2)+(-3+4)+\cdots+(-(n-1)+n)&if $n$ is even\cr
           -1+(2-3)+(4-5)+\cdots+((n-1)-n) &if $n$ is odd\cr}\cr
  &=\cases{k&if $n=2k$\cr -1-k&if $n=2k+1$\cr}\cr
  &=(-1)^n\left\lceil\frac n2\right\rceil\ .\cr}$$
Hence the coefficient of $x^2$ satisfies the recurrence
$$b_{n+1}=b_n+(-1)^{n+1}(n+1)a_n\ ,$$
that is,
$$b_{n+1}=b_n-(n+1)\left\lceil\frac n2\right\rceil\ .$$
It is now pretty easy to calculate by mental arithmetic the values
$$\eqalign{
  b_1,b_2,\ldots,b_{15}&=0,-2,-5,-13,-23,-41,-62,\cr
  &\qquad{}-94,-130,-180,-235,-307,-385,-483,-588.\cr}$$
