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Q.
Find the coefficient of $x^{11}$ in the expansion of $\left(x^{3}-\frac{2}{x^{2}}\right)^{12}$.
Binomial Theorem
Solution:
Let $(r + 1)^{th}$ term contain $x^{11}$.
We have, $T_{r+1} = \,{}^{12}C_{r}\left(x^{3}\right)^{12-r}\left(-\frac{2}{x^{2}}\right)^{r}$
$= \,{}^{12}C_{r}\,x^{36-3r-2r}\,\left(-1\right)^{r}\,2^{r}$
$= \,{}^{12}C_{r}\left(-1\right)^{r}\,2^{r}\,x^{36-5r}$
Now for this to contain $x^{11}$, $36 - 5r = 11$, i.e., $r = 5$
Thus, the coefficient of $x^{11}$ is $^{12}C_{5}\left(-1\right)^{5}2^{5} = -25344$