If the constant term in the expansion of ( 1 + 2 x - 3 x 3 ) ( 3 2 x 2 - 1 3 x ) 9 is p , then 108 p is equal to _______. [2024]

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Solution

Q.
If the constant term in the expansion of $(1 + 2 x - 3 x^{3}) {(\frac{3}{2} x^{2} - \frac{1}{3 x})}^{9}$ is $p,$ then $108 p$ is equal to _______. [2024]

Ans.
(54)

General term of $I = {(\frac{3}{2} x^{2} - \frac{1}{3 x})}^{9}$ is

$T_{r + 1} = {}^{9}{C_{r}} {(\frac{3}{2} x^{2})}^{9 - r} {(\frac{- 1}{3 x})}^{r}$

$= {}^{9}{C_{r}} {(\frac{3}{2})}^{9 - r} \frac{x^{18 - 2 r}}{x^{r}} {(\frac{- 1}{3})}^{r}$

$= {}^{9}{C_{r}} {(\frac{3}{2})}^{9 - r} r^{18 - 3 r} {(\frac{- 1}{3})}^{r}$

Now, $(1 + 2 x - 3 x^{3}) {(\frac{3}{2} x^{2} - \frac{1}{3 x})}^{9}$ will have the constant term $= 1 \times coeff of x^{0} in I + 2 \times coeff of \frac{1}{x} in I - 3 \times coeff of x^{- 3} in I$

$= 1 \times {}^{9}{C_{6}} {(\frac{3}{2})}^{3} {(\frac{- 1}{3})}^{6} + 2 \times 0 - 3 \times {}^{9}{C_{7}} {(\frac{3}{2})}^{2} {(\frac{- 1}{3})}^{7}$

$= \frac{84}{2^{3} \times 3^{3}} + \frac{3 \times 36 \times 1}{2^{2} \times 3^{5}} = \frac{7}{2 \times 3^{2}} + \frac{2}{2 \times 3^{2}}$

$= \frac{9}{2 \times 3^{2}} = \frac{1}{2}$

$∴$ $p = \frac{1}{2} .$ So $108 p = \frac{1}{2} \times 108 = 54$

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