If the constant term, in binomial expansion of (2 x ^ r + 1 x ^ 2 )...

The general term in the binomial expansion of ${\left(a + b\right)}^{n}$ is $CR n {a}^{n - R}{b}^{R} .</p><p>Thus, the general term in the expansion of${\left(2{x}^{r} + \dfrac{1}{{x}^{2}}\right)}^{10} is $= CR 10 {\left(2{x}^{r}\right)}$^{10 - R}{\left(\dfrac{1}{{x}^{2}}\right)}^{R}

= CR 10 \left({2}^{10 - R}{x}^{r\left(10 - R\right)}\right){x}^{- 2R}$</p><p>Given, the constant term in the expansion is$180$</p><p>$⇒ CR 10 \cdot {2}^{10 - R} = 180 \ldots \left(1\right)$</p><p>And$( 10 - R ) r - 2R = 0$</p><p>$⇒ r = \dfrac{2R}{10 - R}$</p><p>$⇒ r = \dfrac{2 ( R - 10 )}{10 - R} + \dfrac{20}{10 - R}$</p><p>$⇒ r = - 2 + \dfrac{20}{10 - R} \ldots \left(2\right)

Since $R$ is positive integer less than or equal to $10 and r$ is also an integer, hence, the value of $R$ is such that, $10 - R$ must divide $20 .</p><p>This is possible, when the value of$10 - R$can be one from the divisors of$20$i.e.$10 - R$can be from$1, 2, 4 or 5 and consequently, $R$ can be $9, 8, 6 or 5$ respectively.

But for $R = 9, 6 \text{ and } 5 equation \left(1\right)$ is not satisfied, hence $R = 8$

as $CR 10 {2}^{10 - R} = 180$

$⇒ r = \dfrac{2 \times 8}{10 - 8} =8 .</p>$