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 :
Putting the values 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 = \[\frac{1}{N}\left[ N+\frac{N(N-1)(N-2)}{3!}+\frac{N(N-1)\,(N-2)(N-3)(N-4)}{5!}+.... \right]\] \[=\frac{1}{N}\left\{ {{\,}^{N}}{{C}_{1}}+{{\,}^{N}}{{C}_{3}}+{{\,}^{N}}{{C}_{5}}+.... \right\}\] \[=\frac{1}{N}\left\{ {{2}^{N-1}} \right\}=\frac{{{2}^{n}}}{n+1}\] \[\{\because N=n+1\}\] Trick: Put n=1, then \[{{S}_{1}}=\frac{^{1}{{C}_{0}}}{1}=\frac{1}{1}=1\] At n=2, \[{{S}_{2}}=\frac{^{2}{{C}_{0}}}{1}+\frac{^{2}{{C}_{2}}}{3}=1+\frac{1}{3}=\frac{4}{3}\] Also (c) \[\Rightarrow \,\,\,{{S}_{1}}=1,{{S}_{2}}=\frac{4}{3}\]
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