A) \[\frac{{{2}^{n-1}}}{n-1}\]
B) \[\frac{{{2}^{n+1}}}{n+3}\]
C) \[\frac{{{2}^{n}}}{n+1}\]
D) \[\frac{{{2}^{n-2}}}{n}\]
Correct Answer: C
Solution :
We know that, \[{{(1+x)}^{n}}={{C}_{0}}+{{C}_{1}}x+{{C}_{2}}{{x}^{2}}+....\] and \[{{(1-x)}^{n}}={{C}_{0}}-{{C}_{1}}x+{{C}_{2}}{{x}^{2}}-....\] On adding, we get \[{{(1+x)}^{n}}+{{(1-x)}^{n}}=2[{{C}_{0}}+{{C}_{2}}{{x}^{2}}+...]\] On integrating between the limits 0 to 1, we get \[\left[ \frac{{{(1+x)}^{n+1}}}{n+1}-\frac{{{(1-x)}^{n+1}}}{n+1} \right]_{0}^{1}\] \[=2\left[ \frac{{{C}_{0}}x}{1}+\frac{{{C}_{2}}{{x}^{3}}}{3}+..... \right]_{0}^{1}\] \[\Rightarrow \] \[\frac{{{2}^{n+1}}}{n+1}=2\left[ \frac{{{C}_{0}}}{1}+\frac{{{C}_{2}}}{3}+..... \right]\] \[\Rightarrow \] \[\frac{{{C}_{0}}}{1}+\frac{{{C}_{2}}}{3}+....=\frac{{{2}^{n}}}{n+1}\]
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