Question

Suppose $\text{det}\left[\begin{array}{cc}\sum _{k=0}^{n}k& \sum _{k=0}^n{}^n{C}_k{k}^2\\ \sum _{k=0}^n{}^n{C}_{k}k& \sum _{k=0}^n{}^n{C}_k{3}^k\end{array}\right]=0$
hold for some positive integer $n.\text{Then}\sum _{k=0}^n\frac{{}^n{C}_k}{k+1}$ equals

• A. 6.20
• B. 6.2

Answer 1

Answer: (A), (B)

$\text{det}\left[\begin{array}{cc}0+1+2+...+n& {}^n{C}_0\left(0^2\right){+}^n{C}_1\left(1^2\right){+}^n{C}_2\left(2^2\right)+...{+}^n{C}_n\left(n^2\right)\\ {}^n{C}_0\left(0\right){+}^n{C}_{1}1{+}^n{C}_{2}2+...{+}^n{C}_{n}n& {}^n{C}_0\left(3^0\right){+}^n{C}_1\left(3^1\right){+}^n{C}_2\left(3^2\right)+...{+}^n{C}_n\left(3^n\right)\end{array}\right]=0$

$\Rightarrow \text{det}\left[\begin{array}{cc}\frac{n(n+1)}{2}& n(n+1)2^{n-2}\\ n{.2}^{n-1}& 4^n\end{array}\right]=0$

$\Rightarrow n(n+1)2^{n-1}\left|\begin{array}{cc}\frac{1}{2}& 2^{n-2}\\ n& 2^{n+1}\end{array}\right|=0$

$\Rightarrow n(n+1)2^{2n-3}\left|\begin{array}{cc}\frac{1}{2}& 1\\ n& 8\end{array}\right|=0$
$\Rightarrow n(n+1)2^{2n-3}(4-n)=0$
Hence, $n=0,-1,4\Rightarrow n=4[\because n>0]$
$\therefore \sum _{k=0}^4\frac{{}^4{C}_k}{k+1}=\frac{{}^4{C}_0}{1}+\frac{{}^4{C}_1}{2}+\frac{{}^4{C}_2}{3}+\frac{{}^4{C}_3}{4}+\frac{{}^4{C}_4}{5}=\frac{31}{5}=6.2$