First slide

Binomial theorem for positive integral Index

Question

$If a_{n} = \sum_{r = 0}^{n} \frac{1}{^{n} C_{r}}, then \sum_{r = 0}^{n} \frac{r}{^{n} C_{r}}$ is equal to

Moderate

A

$(n - 1) a_{n}$
B

${na}_{n}$
C

$\frac{1}{2} {na}_{n}$
D

None of these

Solution

$Let \begin{aligned} b = \sum_{r = 0}^{n} \frac{r}{^{n} C_{r}} = \sum_{r = 0}^{n} \frac{n - (n - r)}{^{n} C_{r}} \\ = n \sum_{r = 0}^{n} \frac{1}{^{n} C_{r}} - \sum_{r = 0}^{n} \frac{n - r}{^{n} C_{r}} \end{aligned}$
$\begin{aligned} = {na}_{n} - \sum_{r = 0}^{n} \frac{n - r}{^{n} C_{n - r}} [∵^{n} C_{r} =^{n} C_{n - r}] \\ = {na}_{n} - b \\ \Rightarrow 2 b = {na}_{n} \Rightarrow b = \frac{n}{2} a_{n} \end{aligned}$

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