The value of ∑k=0n 4n+1Ck+4n+1C2n−k is

A
$2^{4 n} +^{4 n + 1} C_{2 n}$
B
$2^{4 n + 1}$
C
$2^{4 n}$
D
$2^{4 n + 1} +^{4 n + 1} C_{n}$

Solution:

$\begin{aligned} \sum_{k = 0}^{n} (^{4 n + 1} C_{k} +^{4 n + 1} C_{2 n - k}) \\ = \sum_{k = 0}^{2 n} (^{4 n + 1} C_{k}) +^{4 n + 1} C_{2 n} \\ = 2^{4 n} +^{4 n + 1} C_{2 n} \end{aligned}$

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