Find the term independent of x in the expansion of [2x²+1/x]¹²

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$$\begin{gathered} \mathrm{We} \mathrm{have} {\left(2{\mathrm{x}}^2+\frac{1}{\mathrm{x}}\right)}^{12} \\ \mathrm{Let} {\left(\mathrm{r}+1\right)}_{\mathrm{th}} \mathrm{term} \mathrm{be} \mathrm{independent} \mathrm{of} \mathrm{x}. \\ {\mathrm{T}}_{\mathrm{r}+1}{=}^{12}{\mathrm{C}}_{\mathrm{r}} {\left(2{\mathrm{x}}^2\right)}^{\mathrm{r}}{\left(\frac{1}{\mathrm{x}}\right)}^{12-\mathrm{r}} \\ {=}^{12}{\mathrm{C}}_{\mathrm{r}.}{2}^{\mathrm{r}}.\left({\mathrm{x}}^{2\mathrm{r}}.\frac{1}{{\mathrm{x}}^{12-\mathrm{r}}}\right) \\ {=}^{12}{\mathrm{C}}_{\mathrm{r}.}{2}^{\mathrm{r}}.\left({\mathrm{x}}^{2\mathrm{r}-12+\mathrm{r}}\right) \\ {=}^{12}{\mathrm{C}}_{\mathrm{r}.}{2}^{\mathrm{r}}.\left({\mathrm{x}}^{3\mathrm{r}-12}\right) \\ \mathrm{As} \mathrm{above} \mathrm{term} \mathrm{is} \mathrm{independent} \mathrm{of} \mathrm{x}, \mathrm{hence}, \\ 3\mathrm{r}-12=0 \\ \Rightarrow \mathrm{r}=4 \\ \mathrm{Hence} \mathrm{r}+1=5_{\mathrm{th}} \mathrm{term} \mathrm{is} \mathrm{independent} \mathrm{of} \mathrm{x}. \\ {\mathrm{T}}_{4+1}{=}^{12}{\mathrm{C}}_{4.}{2}^4 \\ \mathbf{\Rightarrow }{\mathbf{T}}_{\mathbf{5}}\mathbf{=}\mathbf{16}{\mathbf{.}}^{\mathbf{12}}{\mathbf{C}}_{\mathbf{4}}\mathbf{ }\left(\mathbf{Answer}\right) \end{gathered}$$

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