Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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...

share|improve this question
    
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 at 17:21

3 Answers 3

up vote 3 down vote accepted

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

share|improve this answer

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:

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

share|improve this answer
    
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 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 at 13:28

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

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.