A) \[\frac{{{2}^{n}}+1}{n+1}\]
B) \[\frac{{{2}^{n}}}{n+1}\]
C) \[\frac{{{2}^{n}}+1}{n-1}\]
D) \[\frac{{{2}^{n}}-1}{n+1}\]
Correct Answer: D
Solution :
[d] \[\frac{{{C}_{1}}}{2}+\frac{{{C}_{3}}}{4}+\frac{{{C}_{5}}}{6}+....=\frac{n}{2}+\frac{n(n-1)(n-2)}{4!}\] \[+\frac{n(n-1)(n-2)(n-3)(n-4)}{6!}+....\] \[=\frac{1}{n+1}\left[ \frac{(n+1)n}{2!}+\frac{(n+1)n(n-1)(n-2)}{4!}+ \right.\] \[\left. \frac{(n+1)n(n-1)(n-2)(n-3)(n-4)}{6!}+.... \right]\] \[=\frac{1}{n+1}{{[}^{n+1}}{{C}_{2}}+{{\,}^{n+1}}{{C}_{4}}+{{\,}^{n+1}}{{C}_{6}}+....]\] \[=\frac{1}{n+1}\left[ {{2}^{n+1-1}}{{-}^{n+1}}{{C}_{0}} \right]=\frac{{{2}^{n}}-1}{n+1}\]
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