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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.

Kyan Cheung
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Mike Johnson
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    Welcome to MSE. Your question is phrased as an isolated problem, without any further information or context. This does not match many users' quality standards, so it may attract downvotes, or closed. To prevent that, please [edit] the question. This will help you recognise and resolve the issues. Concretely: please provide context, and include your work and thoughts on the problem. These changes can help in formulating more appropriate answers. – José Carlos Santos Feb 23 '20 at 09:11
  • What do you get if you multiply the first sum by two? Is it similar in any way to the original version? – Marius Feb 23 '20 at 09:11
  • Are you familiar with geometric sums (for the first sum)? The typical way of finding a geometric sum is denoting the sum by $S$, multiplying the sum by the common ratio $r$, and then subtracting them. For the second sum, keep in mind that $\log \left(i^n\right) = n\log(i)$. – KM101 Feb 23 '20 at 09:22

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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 $S_2 = n\log\,(n-1)!$ .

Lucas Henrique
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  • Thank you for your reply! I'm confused on why you multiplied the first sum by 2 and subtracted by 2 – Mike Johnson Feb 23 '20 at 09:26
  • That's the typical way geometric sums are found. Multiplying the the ratio ($2$ in this case) simplifies once you subtract the sums. Check here for the general case. Notice how all the middle terms cancel out, simplifying everything. – KM101 Feb 23 '20 at 09:34
  • The first sum is very “famous”, so I proceeded to sum it in the canonical way. It’s a geometric series: https://en.wikipedia.org/wiki/Geometric_series – Lucas Henrique Feb 23 '20 at 09:35