Binomial Theorem 2 Question 6
9. If $a _n=\sum _{r=0}^{n} \frac{1}{{ }^{n} C _r}$, then $\sum _{r=0}^{n} \frac{r}{{ }^{n} C _r}$ equals
(a) $(n-1) a _n$
(b) $n a _n$
(c) $\frac{1}{2} n a _n$
(d) None of these
(1998, 2M)
Show Answer
Answer:
Correct Answer: 9. (c)
Solution:
- Let $b=\sum _{r=0}^{n} \frac{r}{{ }^{n} C _r}=\sum _{r=0}^{n} \frac{n-(n-r)}{{ }^{n} C _r}$
$$ \begin{aligned} & =n \sum _{r=0}^{n} \frac{1}{{ }^{n} C _r}-\sum _{r=0}^{n} \frac{n-r}{{ }^{n} C _r} \\ & =n a _n-\sum _{r=0}^{n} \frac{n-r}{{ }^{n} C _{n-r}} \quad\left[\because{ }^{n} C _r={ }^{n} C _{n-r}\right] \\ & =n a _n-b \Rightarrow 2 b=n a _n \Rightarrow \quad b=\frac{n}{2} a _n \end{aligned} $$
