Sum up the cos

First use the power reduction formula:

$\cos^4(\theta)=\dfrac{3+4\cos(2\theta)+\cos(4\theta)}{8}$

So, the sum becomes to:

$8S=\displaystyle \sum_{n=1}^{100}\left(3+4\cos\left(\dfrac{2\pi n}{201}\right)+\cos\left(\dfrac{4\pi n}{201}\right)\right)$

Then let $w=e^{2\pi i/201}$ . We know that both $w$ and $w^2$ are primitive 201st roots of unity. Also, $\dfrac{1}{2}(w^n+w^{201-n})=\cos\left(\dfrac{2\pi n}{201}\right)$ and $w+w^2+\cdots+w^{200}=-1$ , so the sum is:

$8S=300+2\displaystyle \sum_{n=1}^{100} (w^n+w^{201-n})+\dfrac{1}{2}\displaystyle \sum_{n=1}^{100} ((w^2)^n+(w^2)^{201-n})$

$8S=300+2(-1)+\dfrac{1}{2}(-1)=\dfrac{595}{2}$

$S=\boxed{\dfrac{595}{16}}$