7.Binomial Theorem
easy

દ્રીપદી  $\left(2 x^{r}+\frac{1}{x^{2}}\right)^{10}$ ના વિસ્તરણમાં જો અચળ પદ $180$ હોય તો $r$ ની કિમંત મેળવો.

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

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