Roots of unity can be used to extract every $n^\mathrm{th}$ term of a series. This method is due to Simpson [1759].

Theorem. Let $\omega=e^{2\pi i/k}$ be a primitive $k^\mathrm{th}$ root of unity. If $f(x)=\sum_{j=0}^{\infty} a_j x^j$ and $n\not\equiv 0\pmod k$ , then $$ \sum_{j=0}^{\infty} a_{kj+n}x^{kj+n}=\frac{1}{k}\sum_{j=0}^{k-1}\omega^{-jn}f(\omega^{j}x $$

Proof. This is a consequence of the fact that $\sum_{j=0}^{k-1}\omega^{jm}=0$ for $m\not\equiv 0\pmod k$ .

Consider the term involving $x^r$ on the right-hand side. It is $$ \frac{1}{k}\sum_{j=0}^{k-1}\omega^{-jn}a_r\omega^{jr}x^r=\frac{1}{k}a_rx^r\sum_{j=0}^{k-1}\omega^{j(r-n) $$ If $r\not\equiv n\pmod k$ , the sum is zero. So the term involving $x^r$ is zero unless $r\equiv n\pmod k$ , in which case it is $a_rx^r$ since each element of the sum is $1$ .

Note that this method is a generalization of the commonly known trick for extracting alternate terms of a series: $$ \frac{1}{2}(f(x)-f(-x) $$ produces the odd terms of $f$ .