First slide

Binomial theorem for positive integral Index

Question

The term independent of $x$ in the expansion of ${(2 x - \frac{3}{x})}^{6}$ is

Easy

A

4320
B

216
C

$- 216$
D

$- 4320$

Solution

Given expansion is ${(2 x - \frac{3}{x})}^{6}$

We have general term in the expansion ${(x + a)}^{n}$

$(∴ T_{r + 1} =^{n} C_{r} x^{n - r} {(a)}^{r}$ be the expansion of ${(x + a)}^{n})$

$T_{r + 1} =^{6} C_{r} {(2 x)}^{6 - r} {(- \frac{3}{x})}^{r}$

$T_{r + 1} =^{6} C_{r} {(x)}^{6 - r} {(2)}^{6 - r} {(- 3)}^{r} {(x^{- 1})}^{r}$

$T_{r + 1} =^{6} C_{r} 2^{6 - r} {(- 3)}^{r} x^{6 - 2 r} .......... (1)$

$x^{6 - 2 r}$ compare with $x^{0}$ because of independent term

$\Rightarrow x^{6 - 2 r} = x^{0}$

Which is void of x if $6 - 2 r = 0$

$∵ r = 3$ this value substitute in Eqn (1)

$∴$ Term without $x$ is $T_{4} =^{6} C_{3} 2^{6 - 3} {(- 3)}^{3}$ $\begin{matrix} (∴^{n} C_{r} = \frac{n!}{(n - r)! r!}) \end{matrix}$

$T_{4} = \frac{6 \times 5 \times 4}{3 \times 2 \times 1} 2^{3} {(- 3)}^{3}$

$T_{4} = - 20 \times 8 \times 27$ $= - 4320$

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