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This question already has an answer here:

Find the generalized sum of $1+2(2)+3(2)^2+4(2^3)+...+n(2^{n-1})$

I rewrote the above sequence into: $\sum_{k=1}^{n} k(2^{k-1})$. The sequence looks like a hybrid of the summation $\sum_{k=1}^{n} k$ and the geometric series $\sum_{k=1}^{n} 2^{k-1}$. However, I'm blocked. I don't know how to precede in finding the generalized formula, if there is a simple one.

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marked as duplicate by tired, Eric Stucky, mfl, user228113, colormegone Jan 31 '16 at 0:53

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

  • $\begingroup$ This is a typical arithmetic- geometric series, let S denote this expression, find 2S and then subtract both expressions $\endgroup$ – Nikunj Jan 30 '16 at 23:43
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It is easier to solve this by first generalizing:

$$ \sum_{k=1}^n kx^{k-1} = \frac d{dx} \sum_{k=1}^n x^k\\ = \frac{d}{dx}\left( \frac{1-x^{n+1}}{1-x}\right)\\ = \frac{-(n+1)x^n(1-x) + (1-x^{n+1})}{(1-x)^2} $$

Now, plugging in $x=2$ gives:

$$ \sum_{k=1}^n k2^{k-1}=(n+1)2^n + (1 - 2^{n+1}). \\ = (n+1)2^n - 2\cdot2^n + 1\\ = (n-1)2^n + 1 $$

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If $S=2^i+2^{i+1}+\cdots+2^{n-1}$, then $2S=2^{i+1}+2^{i+2}+\cdots+2^n$ so subtracting gives $S=2^n-2^i$.

Then $\displaystyle\sum_{k=1}^{n}k2^{k-1}=\sum_{i=0}^{n-1}\sum_{j=i}^{n-1}2^j=\sum_{i=0}^{n-1}(2^n-2^i)=n2^n-(2^n-1)=(n-1)2^n+1$

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