Question:
If the constant term, in binomial expansion of $\left(2 x^{r}+\frac{1}{x^{2}}\right)^{10}$ is 180 , then $r$ is equal to
Solution:
$\left(2 x^{r}+\frac{1}{x^{2}}\right)^{10}$
General term $={ }^{10} \mathrm{C}_{\mathrm{R}}\left(2 \mathrm{x}^{2}\right)^{10-\mathrm{R}} \mathrm{x}^{-2 \mathrm{R}}$
$\Rightarrow 2^{10-\mathrm{R}^{10} \mathrm{C}_{\mathrm{R}}}=180$..........(1)
$\&(10-R) r-2 R=0$
$\mathrm{r}=\frac{2 \mathrm{R}}{10-\mathrm{R}}$
$r=\frac{2(R-10)}{10-R}+\frac{20}{10-R}$
$\Rightarrow \mathrm{r}=-2+\frac{20}{10-\mathrm{R}}$............(2)
$R=8$ or 5 reject equation (1) not satisfied
At $R=8$
