JEE MAIN - Mathematics (2023 - 31st January Evening Shift - No. 17)
The coefficient of $x^{-6}$, in the
expansion of $\left(\frac{4 x}{5}+\frac{5}{2 x^{2}}\right)^{9}$, is
expansion of $\left(\frac{4 x}{5}+\frac{5}{2 x^{2}}\right)^{9}$, is
Answer
5040
Explanation
Coeff of $x^{-6}$ in $\left(\frac{4 x}{5}+\frac{5}{2 x^{2}}\right)^{9}$
$$ \begin{aligned} T_{r+1} & ={ }^{9} C_{r}\left(\frac{4 x}{5}\right)^{9-r}\left(\frac{5}{2 x^{2}}\right)^{r} \\\\ 9-3 r & =-6 \\\\ r & =5 \end{aligned} $$
Coeff of $x^{-6}={ }^{9} C_{5}\left(\frac{4}{5}\right)^{4}\left(\frac{5}{2}\right)^{5}$ $$ =5040 $$
$$ \begin{aligned} T_{r+1} & ={ }^{9} C_{r}\left(\frac{4 x}{5}\right)^{9-r}\left(\frac{5}{2 x^{2}}\right)^{r} \\\\ 9-3 r & =-6 \\\\ r & =5 \end{aligned} $$
Coeff of $x^{-6}={ }^{9} C_{5}\left(\frac{4}{5}\right)^{4}\left(\frac{5}{2}\right)^{5}$ $$ =5040 $$
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