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7.Binomial Theorem
easy
If the constant term, in binomial expansion of $\left(2 x^{r}+\frac{1}{x^{2}}\right)^{10}$ is $180,$ than $r$ is equal to $......$
A
$1$
B
$2$
C
$6$
D
$8$
(JEE MAIN-2021)
Solution
$\left(2 x^{r}+\frac{1}{x^{2}}\right)^{10}$
$\text { General term }={ }^{10} C_{R}\left(2 x^{r}\right)^{10-R} x^{-2 R}$
$\Rightarrow 2^{10-R 10} C_{R}=180 \ldots \ldots . .(1)$
$\,(10-R) r-2 R=0$
$r=\frac{2 R}{10-R}$
$r=\frac{2(R-10)}{10-R}+\frac{20}{10-R}$
$\Rightarrow r=-2+\frac{20}{10-R} \ldots . . . \text { (2) }$
$\mathrm{R}=8$ or $5$ reject equation $(1)$ not satisfied At $R=8$
$2^{10-R 10} \mathrm{C}_{\mathrm{R}}=180 \Rightarrow \mathrm{r}=8$
Standard 11
Mathematics
