jmc

Each of the $x^2$-terms in the expansion of the product is obtained by multiplying the $x$-terms from two of the 15 factors of the product. The coefficient of the $x^2$-term is therefore the sum of the products of each pair of numbers in the set $\{-1,2,-3,\dots ,14,-15\}$. Note that, in general, $$(a_1+a_2+\cdots +a_n{)}^2=a_1^2+a_2^2+\cdots +a_n^2+2\cdot \left(\sum _{1\le i<j\le n}{a}_i{a}_j\right).$$Thus, the coefficient of $x^2$ is $$\begin{array}{rl}\sum _{1\le i<j\le 15}(-1{)}^{i}i(-1{)}^{j}j& =\frac{1}{2}\left({\left(\sum _{k=1}^{15}(-1{)}^{k}k\right)}^2-\sum _{k=1}^{15}{k}^2\right)\\ & =\frac{1}{2}\left((-8{)}^2-\frac{15(15+1)(2\cdot 15+1)}{6}\right)=-588.\end{array}$$$$\text{centerline}\mathbf{OR}$$Let $C$ be the coefficient of $x^2.$ Then $$\begin{array}{rl}f(x)& =(1-x)(1+2x)(1-3x)\cdots (1-15x)\\ & =1+(-1+2-3+\cdots -15)x+C{x}^2+\cdots \\ & =1-8x+C{x}^2+\cdots .\end{array}$$Thus $f(-x)=1+8x+C{x}^2-\cdots$. But $f(-x)=(1+x)(1-2x)(1+3x)\dots (1+15x)$, so $$\begin{array}{rl}f(x)f(-x)& =(1-x^2)(1-4x^2)(1-9x^2)\cdots (1-225x^2)\\ & =1-(1^2+2^2+3^2+\cdots +15^2)x^2+\cdots .\end{array}$$Also $f(x)f(-x)=(1-8x+C{x}^2+\cdots )(1+8x+C{x}^2-\cdots )=1+(2C-64)x^2+\cdots$. Thus $2C-64=-(1^2+2^2+3^3+\cdots +15^2)$, and, as above, $C=\boxed{-588}$.