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Question Number 27882 by das47955@mail.com last updated on 16/Jan/18

$$\left(2\right)\mathbf{Find}\mathbf{the}\mathbf{term}\mathbf{indepen}-$$$$\mathbf{dent}\mathbf{of}\mathbf{x}\mathbf{in}\mathbf{the}\mathbf{expansion}\mathbf{of}$$$${(2{\mathbf{x}}^2+\frac{1}{\mathbf{x}})}^6$$

Answered by Rasheed.Sindhi last updated on 16/Jan/18

$${\mathrm{T}}_{\mathrm{r}+1}=\left(\begin{array}{c}\mathrm{n}\\ \mathrm{r}\end{array}\right){\mathrm{a}}^{\mathrm{n}-\mathrm{r}}{\mathrm{b}}^{\mathrm{r}}$$$${\mathrm{T}}_{\mathrm{r}+1}=\left(\begin{array}{c}6\\ \mathrm{r}\end{array}\right){\left({2\mathrm{x}}^2\right)}^{6-\mathrm{r}}{\left(\frac{1}{\mathrm{x}}\right)}^{\mathrm{r}}$$$$=\left(\begin{array}{c}6\\ \mathrm{r}\end{array}\right){\left(2\right)}^{6-\mathrm{r}}\left({\mathrm{x}}^{12-2\mathrm{r}}\right)\left(\frac{1}{{\mathrm{x}}^{\mathrm{r}}}\right)$$$$=\left(\begin{array}{c}6\\ \mathrm{r}\end{array}\right){\left(2\right)}^{6-\mathrm{r}}\left({\mathrm{x}}^{12-3\mathrm{r}}\right)$$$$\mathrm{If}{\mathrm{T}}_{\mathrm{r}+1}\mathrm{is}\mathrm{free}\mathrm{of}\mathrm{x}$$$$12-3\mathrm{r}=0\Rightarrow \mathrm{r}=4$$$${\mathrm{T}}_{\mathrm{r}+1}={\mathrm{T}}_{4+1}={\mathrm{T}}_5$$$${\mathrm{T}}_5\mathrm{is}\mathrm{free}\mathrm{of}\mathrm{x}$$

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