How can I find a formula for $$\sum_{k=1}^n k2^k$$
I'd also appreciate if someone could indicate some materials about this subject.
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$\sum_{k=0}^{n-1}(k+1)2^{k+1} = 2\sum_{k=0}^{n-1} k2^k + \sum_{k=0}^{n-1} 2^{k+1}$ – mathworker21 Apr 16 '18 at 01:51
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Hint: Use with telescoping series: $(k-1)2^{k+1}-(k-2)2^k=k2^k$ – robjohn Apr 16 '18 at 01:53
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In my answer here I show how to derive the formula for the related sum of $\sum\limits_{k=1}^\infty kp(1-p)^{k-1}=\frac{1}{p}$. With a few tweaks and some algebraic manipulation by taking the difference of two such series it can be modified to your sum. – JMoravitz Apr 16 '18 at 01:57
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This does not seem like a duplicate. That question is about a convergent series and this is about the partial sum of a divergent series. – robjohn Apr 16 '18 at 06:39
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That is, the partial sum is $(n-1)2^{n+1}+2$, and this diverges as $n\to\infty$. – robjohn Apr 16 '18 at 10:42
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Hint:
$$\sum_{k=1}^nx^k=\frac{x(x^n-1)}{x-1}$$
$$\sum_{k=1}^nkx^{k-1}=\frac{d}{dx}\left(\frac{x(x^n-1)}{x-1}\right)$$
$$\sum_{k=1}^nkx^{k}=x\frac{d}{dx}\left(\frac{x(x^n-1)}{x-1}\right)$$
Put $x=2$.
CY Aries
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