$\operatorname{det}\left[\begin{array}{cc}\sum_{k=0}^n k & \sum_{k=0}^n{ }^n C_k k^2 \\ \sum_{k=0}^n{ }^n C_k k & \sum_{k=0}^n{ }^n C_k 3^k\end{array}\right]=0$, holds for some positive integer $n$. Then $\sum_{k=0}^n \frac{{ }^n C_k}{k+1}$ equals
$\frac{n(n+1)}{2} n(n-1) \cdot 2^{n-2}+n \cdot 2^{n-1}$
$n \cdot 2^{n-1} 4^n$
$\frac{n(n+1)}{2} \cdot 4^n-n^2(n-1) \cdot 2^{2 n-3}-n^2 2^{2 n-2}=0$
$\frac{ n ( n +1)}{2}-\frac{ n ^2( n -1)}{8}-\frac{ n ^2}{4}=0$
$n^2-3 n-4=0$
$n =4$
Now $\sum_{k=0}^4 \frac{{ }^4 C_k}{k+1}=\sum_{k=0}^4 \frac{k+1}{5} \cdot{ }^5 C_{k+1} \frac{1}{k+1}$
$=\frac{1}{5} \cdot\left[{ }^5 C _1+{ }^5 C _2+{ }^5 C _3+{ }^5 C _4+{ }^5 C _5\right]=\frac{1}{5}\left[2^5-1\right]=\frac{31}{5}=6.20$
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