Find coefficient of $x^2$ 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: $\lvert C \rvert $
 A: Consider at first a polynomial $a_nx^n+a_{n-1}x^{n-1}+...+a_2x^2+a_1x+a_0$ having $n$ distinct solutions $x_1, x_2,...,x_n$.
We always have $$S_{i}={(-1)^{i}a_{n-i} \over a_n}$$ with $S_{n-i}$ the sum of all products of $i$ distinct numbers chosen among $x_1, x_2,...,x_n$, for $i = 1,...,n$. 
As given $n =15$ and $x_i= {(-1)^{i+1} \over i}$, we get that $S_n={-1 \over {15!}}$. 
Since $a_0=1$, it implies that $a_n=-15!$.
Now $$S_{n-2}= \sum_{i,j=1;i>j}^{15}{S_n \over {x_ix_j}}=S_n \sum_{i,j=1; i>j}^{15}{1 \over {x_ix_j}} = S_n\sum_{i,j=1; i>j}^{15}(-1)^{i+j}ij$$
Then $$\sum_{i,j=1;i>j}^{15}(-1)^{i+j}ij= {1 \over 2}(\sum_{i,j=1}^{15}(-1)^{i+j}ij -\sum_{i=j=1}^{15}(-1)^{i+j}ij)= {1 \over 2}({\sum_{i=1}^{15}(-1)^ii}{\sum_{j=1}^{15}(-1)^jj} - \sum_{i=1}^{15}i^2)={(8^2-1240)\over 2}=-588$$ using the fact that $$\sum_{i=1}^n i^2={n(n+1)(2n+1)\over 6}$$. 
Therefore, $a_2=558$ showing that $|C|=558$.
A: You may consider doing it manually. 
A term of $x^2$ arises from the product of two factors with the $1$s from the rest, so by multiplying all the pairs of coefficients we get our answer. 
Or, our answer is: $$|-1(2-3+4-5 \dots 14-15) + 2(-3+4-5-6\dots -13+14-15) \dots -13(14-15) +14(-15)|$$
$$=|-1(-7) + 2(6-15) - 3(-6) + 4(5-15) -5(-5) \dots -13(-1) +14(-15)|$$
$$=588$$
A: Multiply the monomials one by one and discard the terms of degree above $2$:
$$1-x\\1+2x\to1+ x-2 x^2\\
1 -3 x\to1 -2 x-5 x^2\\
1+ 4 x\to1+ 2 x-13 x^2\\
1 -5 x\to1 -3 x-23 x^2\\
1+ 6 x\to1+ 3 x-41 x^2\\
1 -7 x\to1 -4 x-62 x^2\\
1+ 8 x\to1+ 4 x-94 x^2\\
1 -9 x\to1 -5 x-130 x^2\\
1+ 10 x\to1+ 5 x-180 x^2\\
1 -11 x\to1 -6 x-235 x^2\\
1+ 12 x\to1+ 6 x-307 x^2\\
1 -13 x\to1 -7 x-385 x^2\\
1+ 14 x\to1+ 7 x-483 x^2\\
1 -15 x\to1 -8 x\color{green}{-588} x^2\\$$
Notice that the terms of the first degree follow a simple pattern and that the final result is a sum of pairwise products.
$$0-2\cdot1-3\cdot1-4\cdot2-5\cdot2-\cdots15\cdot7$$

$$(1+ax+bx^2)(1+cx)\to1+(a+c)x+(b+ac)x^2$$
