Question

$ \frac{{{C}_{0}}}{1}+\frac{{{C}_{2}}}{3}+\frac{{{C}_{4}}}{5}+\frac{{{C}_{6}}}{7}+... $ is equal to

• 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} $

Answer 1

Answer: (C)

We know, $ {{(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 $ 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} $