Solution:
Let $b_{k} := a_{9999-k}$ for sake of brevity, so we wish to compute $b_{1} + b_{2} + \cdots + b_{200}$. Let $p_{k}$ be the sum of the $k$-th powers of $\zeta^{m} + \omega^{n}$ over all ordered pairs $(m, n)$ with $0 \leq m < 99$ and $0 \leq n < 101$. Recall that Newton's sums tells us that
$$\begin{aligned} p_{1} + b_{1} & = 0 \\ p_{2} + p_{1} b_{1} + 2 b_{2} & = 0 \\ p_{3} + p_{2} b_{1} + p_{1} b_{2} + 3 b_{3} & = 0 \end{aligned}$$
and in general $k b_{k} + \sum_{j=1}^{k} p_{j} b_{k-j} = 0$. The key idea is that $p_{k}$ is much simpler to compute than $b_{k}$, and we can relate the two with Newton's sums.
The roots of unity filter identity tells us that if $P(s, t)$ is a two-variable polynomial, then $z(P) := \frac{1}{9999} \sum P\left(\zeta^{m}, \omega^{n}\right)$ over all $0 \leq m < 99$ and $0 \leq n < 101$ is exactly the sum of the coefficients of the terms $s^{99a} t^{101b}$. Suppose that $P_{k}(s, t) = (s + t)^{k}$. Then $z\left(P_{k}\right)$ is precisely $p_{k} / 9999$. So one can check that
- if $k \neq 99, 101, 198, 200$, then $(s + t)^{k}$ has no terms of the form $s^{99a} t^{101b}$ and so $p_{k} = 0$.
- if $k = 99, 101, 198$, then $z\left(P_{k}\right) = 1$ and $p_{k} = 9999$.
- if $k = 200$, then $z\left(P_{k}\right) = \binom{200}{99}$ and $p_{k} = 9999 \binom{200}{99}$.
We can now compute $b_{k}$ using Newton's sums identities:
- we have $b_{k} = 0$ for $k \neq 99, 101, 198, 200$.
- since $p_{99} + 99 b_{99} = 0$, we have $b_{99} = -101$;
- since $p_{101} + 101 b_{101} = 0$, we have $b_{101} = -99$;
- since $p_{198} + p_{99} b_{99} + 198 b_{198} = 0$, we have
$$b_{198} = \frac{1}{198}\left(-p_{198} - p_{99} b_{99}\right) = \frac{1}{198}(-9999 + 9999 \cdot 101) = 5050 .$$
- since $p_{200} + p_{101} b_{99} + p_{99} b_{101} + 200 b_{200} = 0$, we have
$$\begin{aligned} b_{200} & = \frac{1}{200}\left(-p_{200} - p_{101} b_{99} - p_{99} b_{101}\right) \\ & = \frac{1}{200}\left(-9999\binom{200}{99} + 9999 \cdot 101 + 9999 \cdot 99\right) \\ & = 9999 - \frac{9999}{200}\binom{200}{99} \end{aligned}$$
Hence, we have
$$\begin{aligned} b_{1} + b_{2} + \cdots + b_{200} & = -101 - 99 + 5050 + 9999 - \frac{9999}{200}\binom{200}{99} \\ & = 14849 - \frac{9999}{200}\binom{200}{99} \end{aligned}$$