G.E.= n2+[n(n-1)(n-2)]/3!4+[n(n-1)(n-2)(n-3)(n-4)]/5!6+… =n2!+n(n-1)(n-2)4!+…=1n+1(n+1)n2!+(n+1)n(n-1)(n-2)4!+… =1n+1C2 n+1+C4 n+1+…=1n+1C0 n+1+C2 n+1+C4 n+1…-C0 n+1 =1n+12n+1-1-1 =2n-1n+1

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$\frac{C_{1}}{2} + \frac{C_{3}}{4} + \frac{C_{5}}{6} + ...$

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$\frac{2^{n+1} -1}{n+1}$

$\frac{2^{n} -1}{n+1}$

$\frac{2^{n}}{n+1}$

$\frac{1}{n+1}$

answer is B.

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Detailed Solution

$\begin{array}{l} G.E.= \frac{n}{2} + \frac{[n (n - 1) (n - 2)] / 3!}{4} + \frac{[n (n - 1) (n - 2) (n - 3) (n - 4)] / 5!}{6} + \dots \\ = \frac{n}{2!} + \frac{n (n - 1) (n - 2)}{4!} + \dots = \frac{1}{n + 1} [\frac{(n + 1) n}{2!} + \frac{(n + 1) n (n - 1) (n - 2)}{4!} + \dots] \\ = \frac{1}{n + 1} [{}^{n + 1}C_{2} + {}^{n + 1}C_{4} + \dots] = \frac{1}{n + 1} [({}^{n + 1}C_{0} + {}^{n + 1}C_{2} + {}^{n + 1}C_{4} \dots) - {}^{n + 1}C_{0}] \\ = \frac{1}{n + 1} [2^{n + 1 - 1} - 1] \\ = \frac{2^{n} - 1}{n + 1} \end{array}$