How do I evaluate $\sum_{i=0}^{n-1}2^i+1$ and $\sum_{i=2}^{n-1} \log i^{n}$?

Question

I'm having trouble with finding the sums of the problem $\sum_{i=0}^{n-1} 2^{i+1}$ and $\sum_{i=2}^{n-1} \log i^{n}$

I've thought it over and don't know where to start with either of these problems.

Answer 1

Tip: call the first sum $S_1$. Then $S_1 = 2S_1 - S_1 = 2^{n+1} - 2$.

Call the second sum $S_2$. Since $\log x^y = y\log x$, $S_2 = \sum\limits_{i=2}^{n-1} n \log i = n \sum\limits_{i=2}^{n-1} \log i$. Now use that $\log x + \log y = \log xy$ to deduce that $S_2 = n\log\,(n-1)!$ .