If $a_{n} = 2 n + 1$ and $C_{r} =^{n} C_{r}$ $= a_{0} C_{0}^{2} + a_{1} C_{1}^{2} + a_{2} C_{2}^{2} + . . . . . . + a_{n} C_{n}^{2}$

(n - 1) (^{2 n} C_{n})

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n (^{2 n} C_{n})

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(n + 1) (^{2 n} C_{n})

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(n + 1) (^{n} C_{n / 2})

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Solution

The correct option is C $(n + 1) (^{2 n} C_{n})$
$S_{n} = a_{0} C_{0}^{2} + a_{1} C_{1}^{2} + a_{2} C_{2}^{2} + . . . . . . + a_{n} C_{n}^{2}$
$S_{n} = a_{n} C_{n}^{2} + a_{n - 1} C_{n - 1}^{2} + a_{n - 2}^{2} + . . . . . . . . . + a_{0} C_{0}^{2}$
$2 S_{n} = (a_{0} + a_{n}) C_{0}^{2} + (a_{1} + a_{n - 1}) C_{1}^{2} + . . . . . . + (a_{n} + a_{0}) C_{n}^{1}$
$= (2 n + 2) (C_{0}^{2} + C_{1}^{2} + C_{2}^{2} + . . . . . . . + C_{n}^{2})$
$∴ S_{n} = (n + 1)^{2 n} C_{n}$
$[∵ a_{0} + a_{n} = a_{1} + a_{n - 1} + . . . . . . . . = 2 n + 2]$

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Similar questions

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