Question

Let $n\in N$ and $\left[x\right]$ denote the greatest integer less than or equal to $x.$ If the sum of $(n+1)$ terms ${}^n{C}_0,3{\cdot }^n{C}_1,5{\cdot }^n{C}_2,7{\cdot }^n{C}_3,\dots$ is equal to $2^{100}\cdot 101,$ then $2\left[\frac{n-1}{2}\right]$ is equal to

Answer 1

$\sum _{r=0}^n(2r+1)\cdot {}^n{C}_r=2^{100}\cdot 101$
$\Rightarrow 2\sum _{r=0}^{n}r\cdot {}^n{C}_r+\sum _{r=0}^n{}^n{C}_r={101\cdot 2}^{100}$
$\Rightarrow 2n\cdot 2^{n-1}+2^n=101\cdot 2^{100}$
$\Rightarrow (n+1)\cdot 2^n=101\cdot 2^{100}$
$\Rightarrow n=100$
$\therefore 2\left[\frac{n-1}{2}\right]=2\left(49\right)=98$