<s>[INST] 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?
 [/INST] 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
$$\eqalign{
  b_1,b_2,\ldots,b_{15}&=0,-2,-5,-13,-23,-41,-62,\cr
  &\qquad{}-94,-130,-180,-235,-307,-385,-483,-588.\cr}$$
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