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Q.
The ratio of the coefficient of $x^{15}$ to the term independent of $x$ in $\left(x^{2}+\frac{2}{x}\right)^{15}$ is
Binomial Theorem
Solution:
Let $T_{r+1}$ be the general term of $\left(x^{2}+\frac{2}{x}\right)^{15}$
$\therefore T_{r+1} =\,{}^{15}C_{r}\left(x^{2}\right)^{15-r}\left(\frac{2}{x}\right)^{r}$
$= \,{}^{15}C_{r}\left(2\right)^{r}\,x^{30-3r}\quad\ldots\left(1\right)$
Now, for the coefficient of term containing $x^{15}$,
$30-3r= 15$, i.e., $r = 5$
Therefore, $^{15}C_{5}\left(2\right)^{5}$ is the coefficient of $x^{15}$ (from $\left(1\right)$)To find the term independent of $x$, put $30-3r=0$
$\Rightarrow r = 10$ Thus $^{15}C_{10 }\,2^{10}$ is the term independent of $x$ (from $\left(1\right)$)
Now the ratio is $\frac{^{15}C_{5}2^{5}}{^{15}C_{10}2^{10}} = \frac{1}{2^{5}} = \frac{1}{32}$