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)

$\frac{{{2}^{n+1}}-1}{n+1}$

Answer 2

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
$\left[ \frac{{{(1+x)}^{n+1}}}{n+1} \right]_{0}^{1}$
$=\left[ {{C}_{0}}x+\frac{{{C}_{1}}{{x}^{2}}}{2}+\frac{{{C}_{2}}{{x}^{3}}}{3}+....+\frac{{{C}_{n}}{{x}^{n+1}}}{n+1} \right]_{0}^{1}$
$\Rightarrow$ $\frac{{{2}^{n+1}}-1}{n+1}={{C}_{0}}+\frac{{{C}_{1}}}{2}+\frac{{{C}_{2}}}{3}+.....+\frac{{{C}_{n}}}{n+1}$