Q56 / 160 · AP EAMCET 2024 (20 May Shift 1) · Mathematics

If $a_n=\sum_{r=0}^n \frac{1}{{ }^n C_r}$ then $\sum_{r=0}^n \frac{r}{{ }^n C_r}=$

  • A
    $(\mathrm{n}-1) \mathrm{a}$
  • B
    n. an
  • Correct answer
    C
    $\frac{n}{2} \cdot a_n$
  • D
    $\mathrm{an}^{+}{ }_1$

Solution

Let $b=\sum_{r=0}^n \frac{r}{{ }^n C_r}$ ....(i) Replace $r$ by $n-r$ $b=\sum_{r=0}^n \frac{n-r}{{ }^n C_{n-r}}$ ....(ii) Adding (i) and (ii) $2 b=\sum_{r=0}^n \frac{r}{{ }^n C_r}+\sum_{r=0}^n \frac{n-r}{{ }^n C_r} \Rightarrow b=\frac{n}{2} \sum_{r=0}^n \frac{1}{{ }^n C_r} \Rightarrow b=\frac{n}{2} a_n$

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

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