Q69 / 170 · Manipal (MET) 2021 · Mathematics
In the usual notation, $\frac{{ }^n C_1}{2}+\frac{{ }^n C_2}{3}+\ldots+\frac{{ }^n C_n}{n+1}$ is equal to
- A$\frac{2^{n+1}-1}{n+1}$
- B$\frac{2^{n+1}-n-1}{n+1}$
- C$\frac{2^{n+1}-n}{n+1}$
- Correct answerDNone of these
Solution
$\begin{aligned} & \frac{{ }^n C_1}{2}+\frac{{ }^n C_2}{3}+\ldots+\frac{{ }^n C_n}{n+1}=\sum_{r=0}^n \frac{{ }^n C_r}{r+1}-{ }^n C_0 \\ = & \sum_{r=0}^n \frac{1}{n+1} \frac{n+1}{r+1}{ }^n C_r-{ }^n C_0 \\ = & \frac{1}{n+1} \sum_{r=0}^n{ }^{n+1} C_{r+1}-{ }^n C_0\end{aligned}$
$\begin{aligned} & =\frac{1}{n+1}\left[\left({ }^{n+1} C_0+{ }^{n+1} C_1+\ldots+{ }^{n+1} C_{n+1}\right)-{ }^{n+1} C_0\right]-C_0 \\ & =\frac{1}{n+1}\left(2^{n+1}-1\right)-1 \\ & =\frac{2^{n+1}-n-1-1}{n+1}=\frac{2^{n+1}-n-2}{n+1}\end{aligned}$
Hence, the correct answer is option: 'D'.