Mathematics - Permutation & Combination Question with Solution | TestHub

$\left(\frac{n+1}{4}\right)\left(2^{n+1}-n-2\right)={\sum }_{k=1}^{n}k\cdot 2^{n+1-k}=S$ where

$$\begin{array}{rlr}& S=1.2^n+2.2^{n-1}+3.2^{n-2}+\dots .+n.2& \end{array}$$

$$\frac{1}{2}S=2^{n-1}+2.2^{n-2}+\dots .+(n-1).2+n$$

Subtracting, $\frac{1}{2}S=\left(2^n+2^{n-1}+\dots ..+2\right)-n=2\left(2^n-1\right)-n$ $\therefore S=2\left(2^{n+1}-2\right)-2n=2\left(2^{n+1}-n-2\right)$

Hence, $\frac{n+1}{4}=2$ (i.e) $n=7$