OMC180(D) - 発想の部分を計算で

　まず $$z^{10} + z^9 + z^8 + z^7 + z^6 + z^5 + 2z^4 + 4z^3 + 8z^2 + 16z + 32 = \left(z + 1\right) \left(z^9 + z^7 + z^5 + 2z^3 + 2z^2 + 6z + 10\right) + 22$$ より $$\sum_{i=1}^{10}\frac1{a_i^6 + a_i^5} = -\frac1{22}\sum_{i=1}^{10} \left(a_i^4 + a_i^2 + 1 + \frac2{a_i^2} + \frac2{a_i^3} + \frac6{a_i^4} + \frac{10}{a_i^5}\right)\mathclose{}.$$ 次に $$\left(z - 1\right) \left(z - 2\right) \left(z^{10} + z^9 + z^8 + z^7 + z^6 + z^5 + 2z^4 + 4z^3 + 8z^2 + 16z + 32\right) = z^{12} - 2z^{11} + z^6 - 64z + 64 \eqqcolon f(z)$$ であって，$f(0) \ne 0$ かつ $$\frac{f(z)}{z^6} = z^6 - 2z^5 + 1 - 2 \left(\frac2z\right)^5 + \left(\frac2z\right)^6$$ より $\rule[-12pt]{0pt}{0pt}f(z) = 0 \;\Longrightarrow\; f\mathopen{}\left(\dfrac2z\right) = 0$．
よって，解と係数の関係と合わせることで $n = 1, \ldots, 5$ のとき $$\sum_{f(z) = 0}z^n = \left(\sum_{f(z) = 0}z\right)^n = 2^n,\qquad \sum_{f(z) = 0}\frac1{z^n} = \frac1{2^n}\sum_{f(z) = 0}\left(\frac2z\right)^n = \frac1{2^n}\sum_{f(z) = 0}z^n = 1,$$ $$\begin{aligned} \therefore&\sum_{i=1}^{10} \left(a_i^4 + a_i^2 + 1 + \frac2{a_i^2} + \frac2{a_i^3} + \frac6{a_i^4} + \frac{10}{a_i^5}\right)\\ &\mathopen{}= \sum_{f(z) = 0} \left(z^4 + z^2 + 1 + \frac2{z^2} + \frac2{z^3} + \frac6{z^4} + \frac{10}{z^5}\right) - \sum_{z = 1, 2} \left(z^4 + z^2 + 1 + \frac2{z^2} + \frac2{z^3} + \frac6{z^4} + \frac{10}{z^5}\right)\\ &\mathopen{}= 10 - \left(1 + 1 + \frac2{2^2} + \frac2{2^3} + \frac6{2^4} + \frac{10}{2^5}\right) = \frac{105}{16}. \end{aligned}$$ これより $$\sum_{i=1}^{10}\frac1{a_i^6 + a_i^5} = -\frac1{22} \times \frac{105}{16} = -\frac{105}{352}$$ であるから，求める値は $\bm{36960}$．