First slide

Binomial theorem for positive integral Index

Question

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

Difficult

A

$\frac{7}{18}$
B

$\frac{5}{8}$
C

$\frac{7}{11}$
D

$\frac{4}{7}$

Solution

$\begin{array}{l} In the expansion {(\frac{3 x^{2}}{2} - \frac{1}{3 x})}^{9} \\ T_{r + 1} = {}^{9}C_{r} {(\frac{3 x^{2}}{2})}^{9 - r \cdot} {(\frac{- 1}{3 x})}^{r} \\ = {}^{9}C_{r} {(\frac{3}{2})}^{9 - r} {(\frac{- 1}{3})}^{r} x^{18 - 3 r} \end{array}$
If 18-3r=0, then r=6.
If 18-3r=-1, then r=19/3.
If 18-3r=-2, then r=20/3.
$\Rightarrow T h e r e a r e n o t e r m s c o n t a i n i n g \frac{1}{x} a n d \frac{1}{x^{2}} i n t h e e x p a n s i o n o f {(\frac{3 x^{2}}{2} - \frac{1}{3 x})}^{9}$
$\begin{array}{l} ∴The term independent of x in the given expansion = {}^{9}C_{6} {(\frac{3}{2})}^{3} {(- \frac{1}{3})}^{6} = {}^{9}C_{3} \frac{1}{6^{3}} \\ = \frac{84}{216} = \frac{7}{18} \end{array}$

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