Be careful when counting how many terms!

We have $(-1)^na_n - (-1)^{n-1}a_{n-1} \; = \; (-1)^{\frac12n(n-1)}n \hspace{2cm} n \ge 2$ and hence that $\begin{aligned} (-1)^na_n & = \; -a_1 + \sum_{m=2}^n (-1)^{\frac12m(m-1)}m \\ a_n & = \; (-1)^{n+1}a_1 + \sum_{m=2}^n (-1)^{\frac12m(m-1) + n}m \end{aligned}$ for $n \ge 2$ and hence that $S_n \; = \; a_1\left(\sum_{k=1}^n (-1)^{k+1}\right) + \sum_{k=2}^n\sum_{m=2}^k (-1)^{\frac12m(m-1)+k}m \; = \; a_1\left(\sum_{k=1}^n (-1)^{k+1}\right) + \sum_{m=2}^n\left(\sum_{k=m}^n(-1)^k\right) (-1)^{\frac12m(m-1)}m$ for $n \ge 2$ . A good deal of cancelling is going on here, so that $S_{2n+1} \; = \; a_1 - \sum_{m=1}^{n}(-1)^{\frac12(2m+1)2m}(2m+1) \; = \; a_1 - \sum_{m=1}^n(-1)^m(2m+1)$ Thus we deduce that $S_{2017} = a_1 - 1008$ , and hence $b = 1-a_1$ . Thus we want to minimize $\frac{2}{a} + \frac{3}{1-a}$ over the range $0 < a < 1$ . This minimum takes place when $a= \frac{\sqrt{2}}{\sqrt{2}+\sqrt{3}}$ , and equals $A = (\sqrt{3}+\sqrt{2})^2 = 5 + 2\sqrt{6}$ . This makes the answer $\boxed{98989}$ .