Question

If $S_{n} = \sum\limits^{n}_{r = 0} \frac{1}{^{n}C_{r}}$ and $t_{n} = \sum\limits^{n}_{r = 0} \frac{r}{^{n}C_{r}}$, then $\frac{t_{n}}{S_{n}}$ is equal to

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

Answer 1

Answer: (A)

$t = \sum\limits^{n}_{r = 0} \frac{1}{^{n}C_{r}} = \sum\limits^{n}_{r = 0} \frac{n-r}{^{n}C_{n-r}} = \sum\limits^{n}_{r = 0} \frac{n-r}{^{n}C_{n-r}}\quad\left(\because\quad^{n}C_{r} = ^{n}C_{n-r}\right)$
$ 2t_{n} = \sum\limits^{n}_{r = 0} \frac{r+n-r}{^{n}C_{r}} = \sum\limits^{n}_{r = 0} \frac{n}{^{n}C_{r}} \Rightarrow t_{n} = \frac{n}{2} \sum\limits^{n}_{r = 0} \frac{1}{^{n}C_{r}} = \frac{n}{2} S_{n} \quad\Rightarrow \quad \frac{t_{n}}{S_{n}} = \frac{n}{2}$