Completely Complex

If $a_1,a_2,...,a_{2n}$ are the roots of the polynomial $f_n(X) \; = \; X^{2n} + X^{2n-1} + \cdots + X + 1$ then $1+a_1,1+a_2,...,1+a_{2n}$ are the roots of the polynomial $g_n(X) \; =\; f_n(X-1) = (X-1)^{2n} + (X-1)^{2n-1} + \cdots + (X-1) + 1$ The last two terms of this polynomial are $\cdots - \big\{2n - (2n-1) + (2n-2) - (2n-3) + \cdots + 2 - 1\big\}X + f_n(-1) \; = \; \cdots -nX + 1$ and hence we deduce that $h_n(X) = X^{2n}g_n(X^{-1})$ has leading terms $h_n(X) \; = \; X^{2n} - nX^{2n-1} + \cdots$ But $h_n(X)$ is the monic polynomial with zeros $\frac{1}{a_1+1},\tfrac{1}{a_2+1},...,\tfrac{1}{a_{2n}+1}$ , and so $\sum_{j=1}^{2n} \frac{1}{a_j + 1} \; = \; n$ Thus $\sum_{j=1}^{2n} \frac{a_j^2}{a_j + 1} \; = \; \sum_{j=1}^n \left(a_j - 1 + \frac{1}{a_j + 1}\right) \; = \; \sum_{j=1}^{2n}a_j - 2n + \sum_{j=1}^{2n}\frac{1}{a_j+1} \; = \; -1 - 2n + n \; = \; -n-1$ making the answer in this case $-1009-1 = \boxed{-1010}$ .