If det $(\begin{matrix} \sum_{k = 0}^{n} k & \sum_{k = 0}^{n}^{n} C_{k} . k^{2} \\ \sum_{k = 0}^{n}^{n} C_{k} & k \sum_{k = 0}^{n}^{n} C_{k} {.3}^{k} \end{matrix}) = 0$ holds for some $n \in N .$ Then $\sum_{k = 0}^{n} \frac{^{n} C_{k}}{k + 1}$ equals

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$\frac{141}{5}$

$\frac{21}{5}$

$\frac{11}{5}$

$\frac{31}{5}$

answer is B.

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Detailed Solution

$|\begin{matrix} \frac{n (n + 1)}{2} & n (n - 1) 2^{n - 2} + n 2^{n - 1} \\ n 2^{n - 1} & 4^{n} \end{matrix}| = 0$

$\Rightarrow n (n + 1) 2^{2 n - 1} = n^{2} 2^{2 n - 2} + n^{2} (n - 1) 2^{2 n - 3} \Rightarrow n^{2} - 3 n - 4 = 0 \Rightarrow n = 4$

$\sum \frac{^{n} C_{k}}{k + 1} = \int_{0}^{1} {(1 + x)}^{n} = \frac{2^{n + 1} - 1}{n + 1} = \frac{31}{5}$

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