Exponential sum simplification I've got this: $$\sum_{k=1}^{n}\frac{1}{2^{2k}}$$Wolfram Alpha already simplified it for me as $\dfrac{2^{2n}-1}{3\cdot2^{2n}}$, but he hasn't told me why.

For any geometric series
$$\sum_{k=1}^n c^k=c\frac{c^n-1}{c-1}$$
so in your case, as Jonathan's comment proposes
$$\sum_{k=1}^n\frac1{2^{2k}}=\sum_{k=1}^n\left(\frac14\right)^k=\frac14\frac{\left(\frac14\right)^n-1}{\frac14-1}=\frac{\frac{1-4^n}{4^n}}{1-4}=\frac{2^{2n}-1}{3\cdot2^{2n}}$$
