Question

If total number of runs scored in $n$ matches is $\left(\frac{n+1}{4}\right)(2^{n+1}-n-2)$ where $n>1,$ and the runs scored in the $k^{\text{th}}$ match are given by $k\cdot 2^{n+1-k}$ where $1\le k\le n,$ then the value of $n$ is

Answer 1

Let $S_n=\sum _{k=1}^{n}k\cdot 2^{n+1-k}=2^{n+1}\sum _{k=1}^{n}k\cdot 2^{-k}$

Let $t_n=\sum _{k=1}^n\frac{k}{2^k}$
$\Rightarrow t_n=\frac{1}{2}+\frac{2}{2^2}+\frac{3}{2^3}+\cdots +\frac{n}{2^n}--\left(1\right)$
$\Rightarrow \frac{t_n}{2}=\frac{1}{2^2}+\frac{2}{2^3}+\cdots +\frac{n}{2^{n+1}}--\left(2\right)$

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

$\therefore S_n=2^{n+1}\times t_n$
$=2^{n+1}\times 2(1-\frac{1}{2^n}-\frac{n}{2^{n+1}})$
$=2[2^{n+1}-2-n]$
On comparing with the expression $\left(\frac{n+1}{4}\right)(2^{n+1}-n-2),$
$\frac{n+1}{4}=2\Rightarrow n=7$