If the constant term in the binomial expansio | vedclass

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(D) The general term in the expansion of $\left(2x^{r} + x^{-2}\right)^{10}$ is given by $T_{k+1} = {}^{10}C_{k} (2x^{r})^{10-k} (x^{-2})^{k}$.For the

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$8$

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(D) The general term in the expansion of $\left(2x^{r} + x^{-2}\right)^{10}$ is given by $T_{k+1} = {}^{10}C_{k} (2x^{r})^{10-k} (x^{-2})^{k}$.For the constant term,the power of $x$ must be $0$,so $r(10-k) - 2k = 0$,which implies $r = \frac{2k}{10-k}$.The constant term is ${}^{10}C_{k} \cdot 2^{10-k} = 180$.Testing integer values for $k$ where $0 \le k \le 10$:If $k = 8$,then ${}^{10}C_{8} \cdot 2^{10-8} = {}^{10}C_{2} \cdot 2^{2} = 45 \cdot 4 = 180$.Substituting $k = 8$ into the equation for $r$:$r = \frac{2(8)}{10-8} = \frac{16}{2} = 8$.

If the constant term in the binomial expansion of $\left(2x^{r} + \frac{1}{x^{2}}\right)^{10}$ is $180$,then $r$ is equal to $......$

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