Question

Suppose
det$\begin{bmatrix} \displaystyle \sum_{k=0}^n k \displaystyle \sum_{k=0}^n{^n}C_kk^2\\ \displaystyle \sum_{k=0}^n{^n}C_kk \displaystyle \sum_{k=0}^n{^n}C_k3^k\end{bmatrix}=0$, holds for some positive integern. Then $ \sum\limits^{n}_{k = 0} \frac{^{n}C_{k}}{k+1}$ equals

Answer 1

$\begin{vmatrix}\frac{n\left(n+1\right)}{2}&n\left(n-1\right).&2^{n-2}&+\, n.2^{n-1}\\ n.2^{n-1}&&4n^{n}&\end{vmatrix}=0$
$\frac{n\left(n+1\right)}{2}.4^{n}-n^{2}\left(n-1\right).2^{2n-3}-n^{2}2^{2n-2}=0$
$\frac{n\left(n+1\right)}{2}-\frac{n^{2}\left(n-1\right)}{8}-\frac{n^{2}}{4}=0$
$n^{2}-3n-4=0$
$n=4$
Now $\displaystyle \sum_{k=0}^4$ $\frac{^{4}C_{k}}{k+1}=$ $\displaystyle \sum_{k=0}^4$ $\frac{k+1}{5}. ^{5}C_{k+1} \frac{1}{k+1}$
$=\frac{1}{5}.\left[^{5}C_{1}+^{5}C_{2}+^{5}C_{3}+^{5}C_{4}+^{5}C_{5}\right]=\frac{1}{5}\left[2^{5}-1\right]=\frac{31}{5}=6.20$