Coefficient of x ^{-6} , in expansion of left(frac{4 x}{5}+frac{5}{2 x^2}right)^9 , is ........ .

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7.Binomial Theorem

hard

The Coefficient of $x ^{-6}$, in the expansion of $\left(\frac{4 x}{5}+\frac{5}{2 x^2}\right)^9$, is $........$.

$5041$

$5042$

$5043$

$5040$

(JEE MAIN-2023)

Solution

$\left(\frac{4 x}{5}+\frac{5}{2 x^2}\right)^9$

Now, $T _{ r +1}={ }^9 C _{ r } \cdot\left(\frac{4 x }{5}\right)^{9- r }\left(\frac{5}{2 x ^2}\right)^{ r }$ $={ }^9 C _{ r } \cdot\left(\frac{4}{5}\right)^{9- r }\left(\frac{5}{2}\right)^{ r } \cdot x ^{9-3 r }$

Coefficient of $x^{-6}$ i.e. $9-3 r=-6 \Rightarrow r=5$

So, Coefficient of $x^{-6}={ }^9 C _5\left(\frac{4}{5}\right)^4 \cdot\left(\frac{5}{2}\right)^0=5040$

Standard 11

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Solution

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