Q43 / 160 · WBJEE 2012 · Mathematics
$\begin{array}{llll}\text { 3. The sum } & \text { of } & \text { the } & \text { series }\end{array}$ $1+\frac{1}{2}{ }^{n} C_{1}+\frac{1}{3}{ }^{n} C_{2}+\quad+\frac{1}{n+1}{ }^{n} C_{n}$ is equal to
- Correct answerA$\frac{2^{n+1}-1}{n+1}$
- B$\frac{3\left(2^{n}-1\right)}{2 n}$
- C$\frac{2^{n}+1}{n+1}$
- D$\frac{2^{n}+1}{2 n}$
Solution
$1+\frac{1}{2}{ }^{n} C_{1}+\frac{1}{3}{ }^{n} C_{2}+\quad+\frac{1}{n+1}{ }^{n} C_{n}$
$=\frac{1}{n+1}\left[(n+1)+\frac{(n+1) n}{21}\right.$
$+\frac{(n+1) n(n-1)}{3 !}+(1]$
$=\frac{1}{n+1}\left[^{n+1} C_{1}+{ }^{n+1} C_{2}+\quad+{ }^{n+1} C_{n+1}\right]$
\(=\frac{2^{n+1}-1}{n+1}\)
Hence, the correct answer is option: 'A'.