Binomial Theorem Question 18

Question 18 - 31 January - Shift 2

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

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Answer: 5040

Solution:

Formula: General term in the expansion, Important Results (Coefficient of $x^{m}$ in the expansion of $\left(a x^{p} + \frac{b}{x^{q}}\right)^{n})$

General term of $(\frac{4 x}{5}+\frac{5}{2 x^{2}})^{9}$ is given by,

$T _{r+1}={ }^{9} C_r \cdot(\frac{4 x}{5})^{9-r}(\frac{5}{2 x^{2}})^{r}$

$={ }^{9} C_r \cdot(\frac{4}{5})^{9-r}(\frac{5}{2})^{r} \cdot x^{9-3 r}$

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

So, Coefficient of $x^{-6}={ }^{9} C_5(\frac{4}{5})^{4} \cdot(\frac{5}{2})^{5}=5040$