Question

The positive integral value of $n$ such that $1\cdot 2^1+2\cdot 2^2+3\cdot 2^3+4\cdot 2^4+\dots +n\cdot 2^n=2+2^{n+5},$ is

Answer 1

Let $S_n=1\cdot 2^1+2\cdot 2^2+3\cdot 2^3+4\cdot 2^4+\dots +n\cdot 2^n\cdots \left(1\right)2.S_n=1\cdot 2^2+2\cdot 2^3+\dots +(n-1)2^n+n\cdot 2^{n+1}\cdots \left(2\right)$

From $\left(1\right)-\left(2\right),$ we get
$\Rightarrow -S_n=(2^1+2^2+2^3+.......+2^n)-n\cdot 2^{n+1}\Rightarrow -S_n=\frac{2(2^n-1)}{2-1}-n\cdot 2^{n+1}\Rightarrow S_n=n\cdot 2^{n+1}-2(2^n-1)\Rightarrow S_n=2+(n-1)2^{n+1}$

Given $2+(n-1)2^{n+1}=2+2^{n+5}$
$\Rightarrow n-1=2^4\therefore n=17$