The sum of the coefficient of x 2 3 and x - 2 5 in the binomial expansion of ( x 2 3 + 1 2 x - 2 5 ) 9 is [2024]

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Solution

Q.
The sum of the coefficient of $x^{\frac{2}{3}}$ and $x^{\frac{- 2}{5}}$ in the binomial expansion of ${(x^{\frac{2}{3}} + \frac{1}{2} x^{\frac{- 2}{5}})}^{9}$ is [2024]

1 $\frac{69}{16}$

2 $\frac{63}{16}$

3 $\frac{21}{4}$

4 $\frac{19}{4}$

Ans.
(3)

$T_{r + 1} = {}^{9}{C_{r}} {(\frac{x^{- 2 / 5}}{2})}^{r} {(x^{2 / 3})}^{9 - r}$

$= {}^{9}{C_{r}} \frac{1}{2^{r}} x^{6 - \frac{2}{3} r - \frac{2}{5} r} = {}^{9}{C_{r}} \frac{1}{2^{r}} x^{6 - \frac{16 r}{15}}$

For coefficient of $x^{2 / 3}, 6 - \frac{16 r}{15} = \frac{2}{3}$

$\Rightarrow 90 - 16 r = 10$

$\Rightarrow r = 5$

For coefficient of $x^{- 2 / 5}, 6 - \frac{16 r}{15} = \frac{- 2}{5}$

$\Rightarrow 90 - 16 r = - 6$

$\Rightarrow r = 6$

Required sum $= {}^{9}{C_{5}} \frac{1}{2^{5}} + {}^{9}{C_{6}} \frac{1}{2^{6}} = \frac{126}{2^{5}} + \frac{84}{2^{6}}$

$= \frac{336}{2^{6}} = \frac{21}{4}$

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