How to do these very abstract summation simplifications?

Question

For instance, I have these questions for an assignment.

$ \sum\limits_{k=0}^{n-1} (n-k)2^{k}$

$ \sum\limits_{k=1}^{n}\frac 1 {4k^{2}-1}$

Answer 1

For sums that include $ka^k$, one trick is to substitute $$ka^k=a\left[\frac{\mathrm{d}}{\mathrm{d}x}x^k\right]_{x=a}.$$

The first sum is \begin{align} \sum_{k=0}^{n-1}(n-k)2^k & = n\sum_{k=0}^{n-1}2^k-\sum_{k=0}^{n-1}k2^k \\ &=n(2^n-1)-2\left[\frac{\mathrm{d}}{\mathrm{d}x}\sum_{k=0}^{n-1} x^k\right]_{x=2} \\ & =n(2^n-1)-2\left[\frac{\mathrm{d}}{\mathrm{d}x}\frac{x^n-1}{x-1}\right]_{x=2} \\ &=n(2^n-1)-2\left[\frac{nx^{n-1}(x-1)-x^n+1}{(x-1)^2}\right]_{x=2} \\ &=2^{n+1}-n-2. \end{align}

For the second sum, refer to this answer. You can write the summands as a sum of simple fractions $$\sum_{k=1}^{n-1}\frac{1}{4k^2-1}=\frac{1}{2}\left[\sum_{k=1}^{n-1}\frac{1}{2k-1}-\sum_{k=1}^{n-1}\frac{1}{2k+1}\right]$$ and note that this is a telescopic cum: the $k$-th term in the first sum is equal to the $(k+1)$-th term in the second sum, so all terms cancel except two: $$\sum_{k=1}^{n-1}\frac{1}{4k^2-1}=\frac{1}{2}\left[1-\frac{1}{2n-1}\right]=\frac{n-1}{2n-1}.$$