If $C_{0}, C_{1}, C_{2}, \dots, C_{11}$ denote the binomial coefficients of
n the expansion of $(1 + x)^{n}$ , then
$C_{0}^{2} + 2 C_{1}^{2} + 3 C_{2}^{2} + \dots + (n + 1) C_{n}^{2} =$

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

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

${(\frac{n}{2} + 1)}^{2 n} C_{n}$

${(\frac{n}{2} + 1)}^{2 n - i} C_{n}$

answer is C.

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Detailed Solution

Let $S = C_{0}^{2} + 2 C_{1}^{2} + 3 C_{2}^{2} + \dots + (n + 1) C_{n}^{2} \dots (i)$

Writings in the reverse order and using $^{n} C_{r} =^{n} C_{n - r},$

we get

$S = (n + 1) {C_{0}}^{2} + n {C_{1}}^{2} + (n - 1) {C_{2}}^{2} + \dots + {C_{n}}^{2}$

Adding (i) and (ii), we get

$\begin{array}{l} 2 S = (n + 2) \{C_{0}^{2} + C_{1}^{2} + \dots . + C_{n}^{2}\} \\ \Rightarrow S = {(\frac{n}{2} + 1)}^{2 n} C_{n} \end{array}$

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