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...
 A: 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
A: $\begin{eqnarray} 
{\bf Hint} \  
  && \color{#0a0}0\cdot 2^2\! &+& \color{#c00}2\cdot 2^2\! &+& 3\cdot 2^3\! &+& 4\cdot 2^4\! &+&\, \cdots &\!+& n\cdot 2^n \\
  &&           &=& \color{#0a0}1\cdot 2^3\! &+& \color{#c00}3\cdot 2^3\! &+& 4\cdot 2^4\! &+&\, \cdots &\!+& n\cdot 2^n \\   
  &&           &&            &=& \color{#0a0}2\cdot 2^4\! &+& \color{#c00}4\cdot 2^4\! &+&\,  \cdots &\!+& n\cdot 2^n \\
  &&           &&            &&               &&   \ddots    &&\ \ \, \vdots &&\\   
  &&           &&            &&               &&             &&\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\! = \color{#0a0}{(n\!-\!2)}\, 2^n\!\! &+& \color{#c00} n\cdot 2^n\\
  &&           &&            &&               &&            &&  &&\!\!\!\!\!\!\!\!\!\!\!\!\!\! =\color{#0a0}{(n\!-\!1)}\, 2^{n+1} 
\end{eqnarray}$
A: 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:
$$S_{n}=4(n-1)2^{n-1}=2^{n+10} \implies n = 2^{9}+1$$
