# Find the partial sum of $\sum n2^{(n-1)}$

Is it possible to find an expression for the partial sum of the series with $n$th term $n2^{(n-1)}$ from $n=1$ to $100$?

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Yes, it is. ${}{}$ –  martini Oct 30 '12 at 8:15
wolframalpha.com/input/?i=\sum_{n%3D1}^{100}n*2^{n-1} –  Julian Kuelshammer Oct 30 '12 at 8:19
Could have sworn I answered a question like this in the past. AH well. Let me get you started. If $x$ is your partial sum, what do you get from $2x-x$? –  Mike Oct 30 '12 at 8:21
As with many interesting integer sequences, if you compute the first few values, you can search for it in Sloane's OEIS; in this case it is A000337. Now knowing a good "guess", one can prove it e.g. by induction. –  Douglas S. Stones Oct 30 '12 at 9:24

\begin{gather}f(x)=\sum\limits_{k=1}^{n}{{x^k}{2^k}}=\sum\limits_{k=1}^{n}{{(2x)^k}}=\dfrac{2x-(2x)^{n+1}}{1-2x} \\ f'(x)=\left(\sum\limits_{k=1}^{n}{{x^k}{2^k}} \right)'=\sum\limits_{k=1}^{n}{kx^{k-1}2^k}=\left(\dfrac{2x-(2x)^{n+1}}{1-2x} \right)' \\ \sum\limits_{k=1}^{n}{k2^{k-1}}=\dfrac{1}{2}f'(1) \end{gather}

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Nice answer. Thanks! The key insight was seeing this as a finite geometric series. I was too used with using the infinite convergent geometric series formula $\frac{a}{1-r}$ and forgot about the general one. –  Legendre Oct 30 '12 at 9:12

$$\sum_{n=1}^m n2^{n-1} =\sum_{n=1}^m \sum_{k=1}^n 2^{n-1} =\sum_{k=1}^m\sum_{n=k}^m 2^{n-1} =\sum_{k=1}^m (2^m-2^{k-1})=m2^m-(2^m-1)$$

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Substitute $x$ for $2$ (and $k$ for $100$) for a moment:

$$\sum_1^k nx^{n-1} = \sum_1^k \dfrac{d}{dx}(x^n) = \dfrac{d}{dx} \sum_1^k x^n = \dfrac{d}{dx}\left(\dfrac{x^{k+1}-x}{x-1}\right)$$

This is all fine because we're just working with finite sums. So evaluate the derivative on the right-hand side and plug in $x=2$.

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The general formula for the sum of first $n$ terms of the series $a , (a+d)r , (a+2d)r^2$ , ... $(a+(n-1)d) r^{n-1}$
is : $$\frac a {1-r} + \frac {dr(1- r^{n-1} ) }{(1-r)^2} - \frac {(a+(n-1)d)r^n} { 1- r}$$
in case of your series $a=1 , d=1 , r = 2 , n = 100$