速通版。
自信構造母函式：
\[\begin{aligned}
f(x) &= (1 + x)(1 + x ^ 2)\cdots(1 + x ^ {2000})\\
&= c_0 + c_1 \times x + c_2 \times x ^ 2 + \cdots
\end{aligned}
\]
取五次單位根 \(\zeta^0, \zeta^1,\cdots,\zeta^4\)。
嘗試計算 \(\sum_{i = 0}^4f(\zeta^i)\)，有：
\[\begin{aligned}
f(\zeta^1) & = [(1 + \zeta)(1 + \zeta^2)(1+\zeta^3)(1+\zeta^4)(1+\zeta^5)]^{400}\\
& = \{[(1+\zeta)(1+\zeta^4)][(1+\zeta^2)(1+\zeta^3)](1+\zeta^5)\}^{400}
\end{aligned}
\]
由於我們有 \(x^5 - 1 = (x - \zeta^0)(x-\zeta^1)\cdots(x-\zeta^4)\)，則取 \(x = -1\) 時 \(2 = (1 + \zeta^0)(1 + \zeta^1)\cdots(1+\zeta^4)\)。
則 \(f(\zeta^1) = 2^{400}\)。易得 \(f(\zeta^2)=f(\zeta^3)=f(\zeta^4)=f(\zeta^1)=2^{400}\)。特殊的，\(f(\zeta^0) = f(1) = 2^{2000}\)。
故 \(\sum_{i = 0}^4f(\zeta^i) = 2^{2000} + 4 \times 2 ^ {400}\)
又得知：
\[\begin{aligned}
\sum_{i = 0}^4f(\zeta^i) &= 5 \times (c_0 + c_5 + \cdots) + (\zeta^0+\zeta^1+\cdots+\zeta^4) \times (c_1 + c_2 + c_3 + c_4 + c_6 + c_7 + \cdots)\\
&= 5\times(c_0+c_5+\cdots)
\end{aligned}
\]
則 \(c_0 + c_5 + \cdots = \frac{1}{5}(2^{2000} + 4 \times 2^{400}) = \frac{1}{5}(2^{2000}+2^{402})\)。