First slide

Binomial theorem for positive integral Index

Question

If ${(1 - x^{2})}^{n} = \sum_{r = 0}^{n} a_{r} x^{r} (1 - x)^{2 n - r}$ , then a_r is equal to

Moderate

A

$^{n} C_{r}$
B

$^{n} C_{r} 3^{r}$
C

$^{2 n} C_{r}$
D

$^{n} C_{r} 2^{r}$

Solution

$(1 - x)^{n} (1 + x)^{n} = \sum_{r = 0}^{n} a_{r} x^{r} (1 - x)^{n} (1 - x)^{n - r}$ $or {(1 + x)}^{n} = \sum_{r = 0}^{n} a_{r} x^{r} {(1 - x)}^{n - r}$
or $(1 - x + 2 x)^{n} = \sum_{r = 0}^{n} a_{r} x^{r} (1 - x)^{n - r}$
or $\sum_{r = 0}^{n}^{n} C_{r} (1 - x)^{n - r} (2 x)^{r} = \sum_{r = 0}^{n} a_{r} x^{r} (1 - x)^{n - r}$
Comparing general term, we get $a_{r} =^{n} C_{r} 2^{r}$ .

Get Instant Solutions

When in doubt download our app. Now available Google Play Store- Doubts App

Recieve an sms with download link

Doubts App

OR

SCAN ME

Doubts App

Doubts App