For each integer $n>1$, let $G(n)$ be the number of solutions to the equation $\cos x=\cos(nx)$ on the interval $[0,\pi]$. What is $\sum_{n=2}^{100}G(n)$?

1. $\cos x=\cos(nx)$ iff $nx=x+2k\pi$ or $nx=-x+2k\pi$, i.e. $x=\dfrac{2k\pi}{n-1}$ or $x=\dfrac{2k\pi}{n+1}$ for an integer $k$.
2. In $[0,\pi]$ the first family has $k=0,1,\dots,\lfloor\frac{n-1}2\rfloor$ and the second has $k=0,1,\dots,\lfloor\frac{n+1}2\rfloor$. A shared solution satisfies $nx\equiv x$ and $nx\equiv-x\pmod{2\pi}$, so $2x\equiv0$ and $x=0$ or $x=\pi$; $x=\pi$ is in both families exactly when $n$ is odd.
3. If $n$ is even this gives $\frac{n}2+\frac{n}2+1-1=n$ solutions; if $n$ is odd it gives $\frac{n+1}2+\frac{n+3}2-2=n$. So $G(n)=n$ for every $n$.
4. $\displaystyle\sum_{n=2}^{100}G(n)=2+3+\cdots+100=\frac{100\cdot101}2-1=\boxed{5049}$.