Computing $\sum_{k=1}^n k 2^k$ by hand

Asked Oct 30 '17 at 14:33

Active Oct 30 '17 at 14:49

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I'm wondering how to determine the closed form for this sum: $$\sum_{k=1}^n k 2^k$$

I'm aware that it is $ 2 (2^n n - 2^n + 1)$ through Wolfram|Alpha, but I wonder how this would be done by hand. Searches for "summations of products" yielded no results, so I'm assuming that there isn't an easy way to determine summations that include products of functions.

How would one approach computing this summation without the help of technology?

asked Oct 30 '17 at 14:33

actinidia

3,365

Compute $\sum_{k=1}^n (k+1) 2^k$ instead. What's the derivative of $\sum_{k=1}^n x^{k+1}$ ? – Gabriel Romon Oct 30 '17 at 14:35
1

https://en.m.wikipedia.org/wiki/Arithmetico–geometric_sequence – lab bhattacharjee Oct 30 '17 at 14:36
See this also. – David Mitra Oct 30 '17 at 14:42
@labbhattacharjee Broken link :P Use proper link formatting [description here](link here) – Simply Beautiful Art Oct 30 '17 at 14:43
You can also transform this into a first order recurrence relation, find an initial value, and compute that – Francisco José Letterio Oct 30 '17 at 15:45

Computing $\sum_{k=1}^n k 2^k$ by hand

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