First slide

Binomial theorem for positive integral Index

Question

The coefficient of the term independent of x in the expansion of $(1 + x + 2 x^{3}) {(\frac{3}{2} x^{2} - \frac{1}{3 x})}^{9}$ is

Moderate

A

$\frac{1}{3}$
B

$\frac{19}{54}$
C

$\frac{17}{54}$
D

$\frac{1}{4}$

Solution

The general term in the expansion of ${(\frac{3}{2} x^{2} - \frac{1}{3 x})}^{9} is$
$\begin{aligned} T_{r + 1} =^{9} C_{r} {(\frac{3}{2} x^{2})}^{9 - r} {(- \frac{1}{3 x})}^{r} \\ =^{9} C_{r} {(\frac{3}{2})}^{9 - r} {(- \frac{1}{3})}^{r} x^{18 - 3 r} \end{aligned} - - - - - - (i)$
Now, the coefficients of the terms $x^{0}, x^{- 1} and x^{- 3} in$
${(\frac{3}{2} x^{2} - \frac{1}{3 x})}^{9} is$
For x^o, 18-3r =0 $\Rightarrow$ + r=6
For x^-1 there exists no integer value of r
For x^-3,18 -3r = -3 $\Rightarrow$ r=7
Now, the coefficient of the term independent of x in the
expansion of $(1 + x + 2 x^{3}) {(\frac{3}{2} x^{2} - \frac{1}{3 x})}^{9}$
$\begin{aligned} = 1 \cdot^{9} C_{6} (- 1)^{6} {(\frac{3}{2})}^{9 - 6} {(\frac{1}{3})}^{6} + 0 \\ + 2 \cdot^{9} C_{7} (- 1)^{7} {(\frac{3}{2})}^{9 - 7} {(\frac{1}{3})}^{7} \end{aligned}$
$\begin{aligned} = \frac{9 \cdot 8 \cdot 7}{1 \cdot 2 \cdot 3} \cdot \frac{3^{3}}{2^{3}} \cdot \frac{1}{3^{6}} + 2 \cdot \frac{9 \cdot 8}{1 \cdot 2} (- 1) \frac{3^{2}}{2^{2}} \cdot \frac{1}{3^{7}} \\ = \frac{7}{18} - \frac{2}{27} = \frac{17}{54} \end{aligned}$

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