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Binomial Theorem Part 3

Section: Mathematics

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Question : 42 of 53

Marks: +1, -0

The sum of the coefficient of

x^{2/3}

x^{-2/5}

in the binomial expansion of

\left(x^{2/3}+\frac{1}{2} x^{-2/5}\right)^9

[9 Apr 2024 Shift 2]

$\frac{21}{4}$
$\frac{69}{16}$
$\frac{63}{16}$
$\frac{19}{4}$

Solution: 👈: Video Solution

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

\; ={}^{9}C_{r} \left(\frac{1}{2}\right)^{r} (r)^{\left(\sigma - \frac{2\pi}{3} \cdot \frac{2r}{5}\right)}

for coefficient of

x^{2/3}

, put

6-\frac{2r}{3}-\frac{2r}{5}=\frac{2}{3}

\Rightarrow r=5

\therefore

Coefficient of

x^{2/3}

is

= {}^{9}C_{5} \left(\frac{1}{5}\right)^{5}

For coefficient of

x^{-2/5}

, put

6-\frac{2r}{3}-\frac{2r}{5}=-\frac{2}{5}

\Rightarrow r=6

Coefficient of

x^{-2/5}

is

{}^{9}C_{6} \left(\frac{1}{2}\right)^{6}

\; \text{Sum}\; = {}^{9}C_{5} \left(\frac{1}{2}\right)^{5} + {}^{9}C_{6} \left(\frac{1}{2}\right)^{6} = \frac{21}{4}

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