For n≥$$2$$, let $$Cr=$$ rn and $$an=$$∑$$r=0n$$ $$1Cr$$

using $$Cr=Cn$$−$$1$$, we get ∑$$r=0n$$ $$rCr=$$∑$$r=0n$$ rCn−$$r=$$∑$$r=0n$$ n−rCr⇒$$2$$∑$$r=0n$$ $$rCr=nan$$⇒∑$$r=0n$$ $$rCr=12nansimilarly$$, ∑$$r=0n$$ n−$$rCr=12nanusing$$, $$rCr=nCr$$−$$1$$ we get∑$$r=0n$$ $$1$$ $$rCr=1n$$∑$$r=1n$$ $$1Cr$$−$$1=1n$$∑$$r=0n$$ $$1Cr=1nan$$−$$1similarly$$ ,∑$$r=0n$$−$$1$$ $$1(n$$−$$r)Cr=1nan$$−$$1$$

For n≥$$2$$, let $$Cr=$$ rn and $$an=$$∑$$r=0n$$ $$1Cr$$

Correct option is 4A-R,B-R,C-Q,D-Q

For n≥$$2$$, let $$Cr=$$ rn and $$an=$$∑$$r=0n$$ $$1Cr$$

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

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

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

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

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

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Binomial theorem for positive integral Index

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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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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. $\frac{1}{2 n} a_{n - 1}$

Binomial theorem for positive integral Index

Unlock the full solution & master the concept.

For n≥2, let Cr= rn and an=∑r=0n 1CrColumn-IColumn-IIA. ∑r=0n rCrP. nanB. ∑r=0n n−rCrQ. 1nan−1C. ∑r=0′n 1 rCrR. 12nanD. ∑r=0n−1 1(n−r)CrS. 12nan−1

using Cr=Cn−1, we get ∑r=0n rCr=∑r=0n rCn−r=∑r=0n n−rCr⇒2∑r=0n rCr=nan⇒∑r=0n rCr=12nansimilarly, ∑r=0n n−rCr=12nanusing, rCr=nCr−1 we get∑r=0n 1 rCr=1n∑r=1n 1Cr−1=1n∑r=0n 1Cr=1nan−1similarly ,∑r=0n−1 1(n−r)Cr=1nan−1

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