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I'm trying to calculate the value of

$$2^0\cdot0 + 2^1\cdot1 + 2^2\cdot2 + .... 2^n\cdot n$$

I figured this would be summation $2^i \cdot i$ from $i = 0$ to $n$. But iI'm unable to calculate its value.

I have tried searching online but haven't been able to find a formula or any property that could simplify it (maybe there was which I might not have understood)

P.S.This is not a homework question, I need this value to prove a theorem.

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  • $\begingroup$ Is this $2^{2.2}$ or is it $2^2\times2$? $\endgroup$
    – user122283
    Commented Apr 24, 2014 at 22:54
  • $\begingroup$ the second one 2^2 x 2 $\endgroup$
    – otaku
    Commented Apr 24, 2014 at 22:55

3 Answers 3

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Hint: Note that $nx^n=x(x^n)'$, so your series is the derivative of a geometric series evaluated at $x=2$.

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WolframAlpha claims that $$ \sum_{n=0}^{N}2^{n}n=2+2^{N+1}(N-1) $$ We can verify this using induction. Clearly it holds for $N=0$. Now assuming the formula holds for $N$, we have $$ \sum_{n=0}^{N+1}2^{n}n=2+2^{N+1}(N-1)+2^{N+1}(N+1)=2+2^{n+1}(2*N)=2+2^{N+2}(N+1-1) $$ And so the formula holds for all $N\in\{0,1,2,\ldots\}$.

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$$\sum^m_{n=0}2^n\times n=2^{m+1}m-2^{m+1}+2$$ The limit as $m\to\infty$ is $\infty$. Or, using the limit test, we directly see that $\sum^\infty_{n=0}2^n\times n$ diverges.

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  • $\begingroup$ If that's all you're after, it's easier than that's every term is greater than $1$, so of course the series diverges. $\endgroup$
    – MPW
    Commented Apr 24, 2014 at 23:20
  • $\begingroup$ @MPW Well, yes - that's what the limit test implies. $\endgroup$
    – user122283
    Commented Apr 24, 2014 at 23:21

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