If total number of runs scored in $n$ matches is $\left(\frac{n+1}{4}\right)\left(2^{n+1}-n-2\right)

Question

If total number of runs scored in $n$ matches is $\left(\frac{n+1}{4}ight)\left(2^{n+1}-n-2ight)$ where $n>1$, and the runs scored in the $k^{	ext {th }}$ match are given by $k \cdot 2^{n+1-k}$, where $1 \leq k \leq n$. Find $n$.\

Professor PC Thomas Classes · Answer

$S_{n}=\sum_{k=1}^{n} k 2^{n+1-k}$\ $=1 \cdot 2^{n}+2 \cdot 2^{n-1}+3 \cdot 2^{n-2}$ $$\begin{equation*}+\ldots .+(n-1) 2^{2}+n(2) 	ag{1} \end{equation*}$$ $\frac{1}{2} S_{n}=1 \cdot 2^{n-1}+2 \cdot 2^{n-2}+3 \cdot 2^{n-3}$ $$\begin{equation*}+\ldots .+(n-1) 2+n 	ag{2} \end{equation*}$$ Subtracting (2) from (1), we get\ $\frac{1}{2} S_{n}=1 \cdot 2^{n}+1 \cdot 2^{n-1}+1 \cdot 2^{n-2}+\ldots .+2-n$ $$ =\frac{2\left(2^{n}-1ight)}{2-1}-n $$ $\Rightarrow S_{n}=2\left(2^{n+1}-2-night)$\ But $S_{n}=\left(\frac{n+1}{4}ight)\left(2^{n+1}-2-night) \quad$ (given)\ $	herefore \quad \frac{n+1}{4}\left(2^{n+1}-2-night)=2\left(2^{n+1}-2-night)$\ $\Rightarrow \frac{n+1}{4}=2 \Rightarrow n=7$\