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Can you help me prove this two sigmas:

$$\sum_{k=2}^n 2^{1-k} = 2^{-n}(2^n - 2) $$

and this

$$\sum_{k=2}^n k2^{1-k} = 3-2^{1-n}(n+2) $$

Is there any common way to solve this kind of tasks? Thank you very much for help.

Jose Avilez
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Blaze Blaut
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  • Multiply the first one by $1-2^{-1}$ and the second by $1-1+2^{-2}$, but don't cancel anything until you have first distributed the multiplication. – plop Jun 07 '21 at 18:34
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    First one: geometric sum. Second one: https://math.stackexchange.com/questions/119636/formula-for-calculating-sum-n-0mnrn (and many other similar questions on this site). – Hans Lundmark Jun 07 '21 at 18:38

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