Find coefficient of $x^2$

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 $|C|=558$.