Question

Suppose
det$\left[\begin{array}{c}\sum _{k=0}^{n}k\sum _{k=0}^n{}^n{C}_k{k}^2\\ \sum _{k=0}^n{}^n{C}_{k}k\sum _{k=0}^n{}^n{C}_k{3}^k\end{array}\right]=0$, holds for some positive integern. Then $\sum _{k=0}^n\frac{{}^n{C}_k}{k+1}$ equals

Answer 1

Answer: 6.2

$\left|\begin{array}{cccc}\frac{n\left(n+1\right)}{2}& n\left(n-1\right).& 2^{n-2}& +n.2^{n-1}\\ n.2^{n-1}& & 4n^n& \end{array}\right|=0$
$\frac{n\left(n+1\right)}{2}.4^n-n^2\left(n-1\right).2^{2n-3}-n^2{2}^{2n-2}=0$
$\frac{n\left(n+1\right)}{2}-\frac{n^2\left(n-1\right)}{8}-\frac{n^2}{4}=0$
$n^2-3n-4=0$
$n=4$
Now $\sum _{k=0}^4$ $\frac{{}^4{C}_k}{k+1}=$ $\sum _{k=0}^4$ $\frac{k+1}{5}{.}^5{C}_{k+1}\frac{1}{k+1}$
$=\frac{1}{5}.\left[{}^5{C}_1{+}^5{C}_2{+}^5{C}_3{+}^5{C}_4{+}^5{C}_5\right]=\frac{1}{5}\left[2^5-1\right]=\frac{31}{5}=6.20$