First slide

Binomial theorem for positive integral Index

Question

$T h e c o e f f i c i e n t o f t h e t e r m i n d e p e n d e n t o f x i n t h e e x p a n s i o n$ $of (1 + x + 2 x^{3}) {(\frac{3}{2} x^{2} - \frac{1}{3 x})}^{9} is$

Moderate

A

$\frac{1}{3}$
B

$\frac{19}{54}$
C

$\frac{17}{54}$
D

$\frac{1}{4}$

Solution

$\begin{array}{l} {The r}^{th} term in the expansion of {(\frac{3}{2} x^{2} - \frac{1}{3 x})}^{9} is T_{r + 1} =^{9} C_{r} {(\frac{3}{2} x^{2})}^{9 - r} {(- \frac{1}{3 x})}^{r} =^{9} C_{r} {(\frac{3}{2})}^{9 - r} {(- \frac{1}{3})}^{r} x^{18 - 3 r} \\ The coefficient of the term independent of x in the expansion of (1 + x + 2 x^{3}) is {(\frac{3}{2} x^{2} - \frac{1}{3 x})}^{9} \\ = Sum of the coefficient of the terms x^{0}, x^{- 1} and x^{- 3} in (\frac{3}{2} x^{2} - \frac{1}{3 x}) \\ Hence the term indepndent of x is = 1 \times^{9} C_{6} {(\frac{3}{2})}^{9 - 6} {(\frac{1}{3})}^{6} + 2 \times^{9} C_{7} {(\frac{3}{2})}^{9 - 7} {(\frac{1}{3})}^{7} \\ = \frac{9.8.7}{1.2.3} \cdot \frac{3^{3}}{2^{3}} \cdot \frac{1}{3^{6}} + 2 \frac{9.8}{1.2} (- 1) \frac{3^{2}}{2^{2}} \cdot \frac{1}{3^{7}} = \frac{7}{18} - \frac{2}{27} = \frac{17}{54} \end{array}$

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