Question

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

Answer 1

Finding the value of $\sum _{k=0}^n\frac{{}^n{C}_k}{k+1}$

Step 1: Consider the given determinant as,

$\left|\begin{array}{cc}\underset{k=0}{\overset{n}{\sum k}}& \sum _{k=0}^n{}^n{C}_k{k}^2\\ \sum _{k=0}^n{}^n{C}_{k}k& \sum _{k=0}^n{}^n{C}_k{3}^k\end{array}\right|=0$

From the standard formula, we can write the above determinant as,

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

Step 2: Simplify the above determinant as,

$$\begin{gathered} \Rightarrow \frac{n\left(n+1\right)}{2}.4^n-n^2\left(n+1\right)2^{2n-3}=0 \\ \Rightarrow \frac{4^n}{2}-n\frac{4^{n-1}}{2}=0 \\ \Rightarrow \frac{4^n}{2}=n\frac{4^{n-1}}{2} \\ \Rightarrow n=\frac{2}{4^{n-1}}\times \frac{4^n}{2} \\ \Rightarrow n=4^{n-n+1} \\ \Rightarrow n=4 \end{gathered}$$

Step 3 : Find the value of $\sum _{k=0}^n\frac{{}^n{C}_k}{k+1}$

$$\begin{gathered} \sum _{k=0}^n\frac{{}^n{C}_k}{k+1}=\sum _{k=0}^4\frac{{}^4{C}_k}{k+1} \\ =\frac{1}{5}\sum _{k=0}^4{}^5{C}_{k+1} \\ =\frac{1}{5}\left(2^5-1\right) \\ =\frac{1}{5}\left(32-1\right) \\ =\frac{31}{5} \\ =6.20 \end{gathered}$$

Therefore, the value of $\sum _{k=0}^n\frac{{}^n{C}_k}{k+1}=6.20$