$\cos^2(\frac{\pi}{101})+\cos^2(\frac{2\pi}{101})+\cos^2(\frac{3\pi}{101})+...+\cos^2(\frac{100\pi}{101})=?$

$$\cos\left(\frac{k\pi}{101}\right)= \frac{1}{2} \left(e^{i\frac{k\pi}{101}}+e^{-i\frac{k\pi}{101}} \right) \\ \cos^2\left(\frac{k\pi}{101}\right)= \frac{1}{4} \left(e^{2i\frac{k\pi}{101}}+e^{-2i\frac{k\pi}{101}} +2\right) \\ \sum_{k=1}^{100}\cos^2\left(\frac{k\pi}{101}\right)= \frac{1}{4} \sum_{k=1}^{100}\left(e^{2i\frac{k\pi}{101}}+e^{-2i\frac{k\pi}{101}} +2\right) $$

Now, $$1+\sum_{k=1}^{100}e^{2i\frac{k\pi}{101}}=\sum_{k=0}^{100}\left(e^{2i\frac{\pi}{101}}\right)^k=\frac{1-(e^{2i\frac{\pi}{101}})^{101}}{1-e^{2i\frac{\pi}{101}}}=0 \\ 1+\sum_{k=1}^{100}e^{-2i\frac{k\pi}{101}}=\sum_{k=0}^{100}\left(e^{-2i\frac{\pi}{101}}\right)^k=\frac{1-(e^{-2i\frac{\pi}{101}})^{101}}{1-e^{-2i\frac{\pi}{101}}}=0 $$

Therefore $$\sum_{k=1}^{100}\cos^2\left(\frac{k\pi}{101}\right)= \frac{1}{4} \left(-1-1+200\right) $$

HINT:

Note that

$$\cos^2(x)=\frac{1+\cos(2x)}{2}$$

Then, the problem boils down to evaluating (See This Answer)

$$\sum_{k=1}^{100}\cos(2k\pi/101)=\text{Re}\left(\sum_{k=1}^{100}\left(e^{i2\pi/101}\right)^k\right)$$

The sum $\sum_{k=1}^{100}\cos(2k\pi/101)$ can also be easily evaluated by multiplying by $\frac{\sin(2\pi/101)}{\sin(2\pi/101)}$ and creative telescoping.

The final answer is $\frac12 (100-1)=\frac{99}{2}$.

HINT

Use the fact that $$\cos^2(x) = \dfrac{1}{2}+\dfrac{\cos(2x)}{2}$$

Thus, the result is equal to

$$50 + \dfrac{1}{2}\sum\limits_{k=1}^{100} \cos\left(\dfrac{2\pi k}{101}\right)$$

And the last sum can be proven to be $-1$ using this result: How to prove $\sum_{k=1}^n \cos(\frac{2 \pi k}{n}) = 0$ for any n>1? .

So the answer is $49.5$.