Question

If $ {{C}_{0}},{{C}_{1}},{{C}_{2}},.....{{C}_{n}} $ denotes the binomial coefficients in the expansion of $ {{(1+x)}^{n}}, $ then $ {{C}_{0}}+\frac{{{C}_{1}}}{2}+\frac{{{C}_{2}}}{3}+....+\frac{{{C}_{n}}}{n+1} $ is equal to

• A. $ \frac{{{2}^{n+1}}-1}{n+1} $
• B. $ \frac{{{2}^{n}}-1}{n} $
• C. $ \frac{{{2}^{n-1}}-1}{n-1} $
• D. $ \frac{{{2}^{n+1}}-1}{n+2} $

Answer 1

Answer: (A)

We know, $ {{(1+x)}^{n}}={{C}_{0}}+{{C}_{1}}x+{{C}_{2}}{{x}^{2}}+....+{{C}_{n}}{{x}^{n}} $
On integrating both sides 0 to 1, we get
$ \left[ \frac{{{(1+x)}^{n+1}}}{n+1} \right]_{0}^{1} $
$ =\left[ {{C}_{0}}x+\frac{{{C}_{1}}{{x}^{2}}}{2}+\frac{{{C}_{2}}{{x}^{3}}}{3}+....+\frac{{{C}_{n}}{{x}^{n+1}}}{n+1} \right]_{0}^{1} $
$ \Rightarrow $ $ \frac{{{2}^{n+1}}-1}{n+1}={{C}_{0}}+\frac{{{C}_{1}}}{2}+\frac{{{C}_{2}}}{3}+.....+\frac{{{C}_{n}}}{n+1} $