Q92 / 225 · JEE Main 2004 · Mathematics

If $S_n=\sum_{r=0}^n \frac{1}{{ }^n C_r}$ and $t_n=\sum_{r=0}^n \frac{r}{{ }^n C_r}$, then $\frac{t_n}{S_n}$ is equal to

  • Correct answer
    A
    $\frac{1}{2} \mathrm{n}$
  • B
    $\frac{1}{2} n-1$
  • C
    $n-1$
  • D
    $\frac{2 n-1}{2}$

Solution

$t=\sum_{r=0}^n \frac{r}{{ }^n C_r}=\sum_{r=0}^n \frac{n-r}{{ }^n C_{n-r}}=\sum_{r=0}^n \frac{n-r}{{ }^n C_r}\left(\because{ }^n C_r={ }^n C_{n-r}\right)$ $2 t_n=\sum_{r=0}^n \frac{r+n-r}{{ }^n C_r}=\sum_{r=0}^n \frac{n}{{ }^n C_r} \Rightarrow t_n=\frac{n}{2} \sum_{r=0}^n \frac{1}{{ }^n C_r}=\frac{n}{2} S_n \Rightarrow \frac{t_n}{S_n}=\frac{n}{2}$

Hence, the correct answer is option: 'A'.

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