First slide

Binomial theorem for positive integral Index

Question

For n $\geq$ 2, let $C_{r} = (_{r}^{n})$ and $a_{n} = {\sum_{r = 0}^{n}}_{} \frac{1}{C_{r}}$
Column-I Column-II
A. $\sum_{r = 0}^{n} \frac{r}{C_{r}}$ P. ${na}_{n}$
B. $\sum_{r = 0}^{n} \frac{n - r}{C_{r}}$ Q. $\frac{1}{n} a_{n - 1}$
C. $\sum_{r = 0}^{' n} \frac{1}{^{r} C_{r}}$ R. $\frac{1}{2} {na}_{n}$
D. $\sum_{r = 0}^{n - 1} \frac{1}{(n - r) C_{r}}$ S. $\frac{1}{2 n} a_{n - 1}$

Moderate

A

A-P,B-R,C-Q,D-S
B

A-R,B-S,C-R,D-Q
C

A-R,B-S,C-Q,D-R
D

A-R,B-R,C-Q,D-Q

Solution

using $C_{r} = C_{n - 1}, we get \sum_{r = 0}^{n} \frac{r}{C_{r}} = \sum_{r = 0}^{n} \frac{r}{C_{n - r}} = \sum_{r = 0}^{n} \frac{n - r}{C_{r}}$
$\Rightarrow 2 \sum_{r = 0}^{n} \frac{r}{C_{r}} = {na}_{n} \Rightarrow \sum_{r = 0}^{n} \frac{r}{C_{r}} = \frac{1}{2} {na}_{n}$
similarly, $\sum_{r = 0}^{n} \frac{n - r}{C_{r}} = \frac{1}{2} {na}_{n}$
using, ${rC}_{r} = {nC}_{r - 1}$ we get
$\sum_{r = 0}^{n} \frac{1}{^{r} C_{r}} = \frac{1}{n} \sum_{r = 1}^{n} \frac{1}{C_{r - 1}} = \frac{1}{n} \sum_{r = 0}^{n} \frac{1}{C_{r}} = \frac{1}{n} a_{n - 1}$
similarly , $\sum_{r = 0}^{n - 1} \frac{1}{(n - r) C_{r}} = \frac{1}{n} a_{n - 1}$

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