# Elementary problems

Find $n$ such that $n$ is a positive integer satisfying the following equation.

$$2(2^2)+3(2^3)+4(2^4)+\ldots+n(2^n)=2^{n+10}$$

Can anybody help ? I can't believe this is an elementary school problem...

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Hint $\$ Add the first two terms of the series, then add the first two terms of the result, etc, etc and you will quickly infer a closed form. See my answer for further details. In particular, you do not need to know the arithmetico-geometric formula, and this method yields simpler arithmetic. – Bill Dubuque Apr 26 '14 at 17:21

Let $\displaystyle S=2\cdot2^2+3\cdot2^3+\cdots+(n-1)2^{n-1}+n\cdot2^n$

$$\implies2S=2\cdot2^3+3\cdot2^4+\cdots+(n-1)2^{n}+n\cdot2^{n+1}$$

$$S-2S=2\cdot2^2+(3-2)2^3+(4-3)2^4+\{n-(n-1)\}2^n-n\cdot2^{n+1}$$

$$\implies -S=2\cdot2^2-n\cdot2^{n+1}+\underbrace{(2^3+2^4+\cdots+2^n)}_{\text{Geomteric Series}}$$

$$\implies-S=2\cdot2^2-n\cdot2^{n+1}+2^3\cdot\frac{(2^{n-2}-1)}{(2-1)}$$

$$\implies S=(n-1)2^{n+1}$$ which needs to be $\displaystyle2^{n+10}\implies n-1=2^9$

Reference : Arithmetico-geometric sequence

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$\begin{eqnarray} {\bf Hint} \ && \color{#0a0}0\cdot 2^2\! &+& \color{#c00}2\cdot 2^2\! &+& 3\cdot 2^3\! &+& 4\cdot 2^4\! &+&\, \cdots &\!+& n\cdot 2^n \\ && &=& \color{#0a0}1\cdot 2^3\! &+& \color{#c00}3\cdot 2^3\! &+& 4\cdot 2^4\! &+&\, \cdots &\!+& n\cdot 2^n \\ && && &=& \color{#0a0}2\cdot 2^4\! &+& \color{#c00}4\cdot 2^4\! &+&\, \cdots &\!+& n\cdot 2^n \\ && && && && \ddots &&\ \ \, \vdots &&\\ && && && && &&\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\! = \color{#0a0}{(n\!-\!2)}\, 2^n\!\! &+& \color{#c00} n\cdot 2^n\\ && && && && && &&\!\!\!\!\!\!\!\!\!\!\!\!\!\! =\color{#0a0}{(n\!-\!1)}\, 2^{n+1} \end{eqnarray}$

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Note that we have an arithmetico-geometric series:

$$S_{n+1}=\sum_{k=1}^{n}2^{2}(1+k)2^{k-1}=4\sum_{k=1}^{n}(2+(k-1))2^{k-1}$$

These have closed form as follows:

$$S_{n+1}=4\left[\frac{2-(2+(n-1))2^{n}}{1-2}+\frac{2(1-2^{n-1})}{(1-2)^{2}}\right]=4\left[-2+2^{n}(n+1)+2-2^{n}\right]=4n2^{n}$$

We therefore want to solve:

$$S_{n}=4(n-1)2^{n-1}=2^{n+10} \implies n = 2^{9}+1$$

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I think that the answer is $2^9+1=513$ and that we have to solve $2^{n+1} (n-1)=2^{n+10}$ – Claude Leibovici Apr 26 '14 at 13:27
@ClaudeLeibovici We are, my expression for $S_{n}$ should be $S_{n+1}$ giving us $4(n-1)2^{n-1}=2^{n+10}$ to solve. – Shaktal Apr 26 '14 at 13:28