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} $
