A) \[\frac{{{2}^{n+1}}}{n+1}\]
B) \[\frac{{{2}^{n+1}}-1}{n+1}\]
C) \[\frac{{{2}^{n}}}{n+1}\]
D) None of these
Correct Answer: C
Solution :
[c] Putting the value of \[{{C}_{0}},{{C}_{2}}{{C}_{4}}......,\] we get \[=1+\frac{n(n-1)}{3.2!}+\frac{n(n-1)(n-2)(n-3)}{5.4!}+.....\] \[=\frac{1}{n+1}\] \[\left[ (n+1)+\frac{(n+1)n(n-1)}{3!}+\frac{(n+1)n(n-1)(n-2)(n-3)}{5!}+.. \right]\]put \[n+1=N\] \[=\frac{1}{N}\left[ N+\frac{N(N-1)(N-2)}{3!}+\frac{N(N-1)(N-2)(N-3)(N-4)}{5!}+.. \right]\]\[=\frac{1}{N}\left\{ ^{N}{{C}_{1}}+{{\,}^{N}}{{C}_{3}}+{{\,}^{N}}{{C}_{5}}+..... \right\}\] \[=\frac{1}{N}\left\{ {{2}^{N-1}} \right\}=\frac{{{2}^{n}}}{n+1}.\left\{ \because \,N=n+1 \right\}\]
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