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Question Number 27885 by das47955@mail.com last updated on 16/Jan/18

$$\left(4\right)\mathbf{Find}\mathbf{the}\mathbf{term}\mathbf{indepen}-$$$$\mathbf{dent}\mathbf{of}\mathbf{x}\mathbf{in}\mathbf{the}\mathbf{expansion}\mathbf{of}$$$${({\mathbf{x}}^2-2+\frac{1}{{\mathbf{x}}^2})}^6$$

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

$${\mathbf{x}}^2-2+\frac{1}{{\mathbf{x}}^2}={(\mathrm{x}-\frac{1}{\mathrm{x}})}^2$$$${({\mathbf{x}}^2-2+\frac{1}{{\mathbf{x}}^2})}^6={\left\{{(\mathrm{x}-\frac{1}{\mathrm{x}})}^2\right\}}^6$$$$={(\mathrm{x}-\frac{1}{\mathrm{x}})}^{12}$$$${\mathrm{T}}_{\mathrm{r}+1}=\left(\begin{array}{c}\mathrm{n}\\ \mathrm{r}\end{array}\right){\mathrm{a}}^{\mathrm{n}-\mathrm{r}}{\mathrm{b}}^{\mathrm{r}}$$$${\mathrm{T}}_{\mathrm{r}+1}=\left(\begin{array}{c}12\\ \mathrm{r}\end{array}\right){\left(\mathrm{x}\right)}^{12-\mathrm{r}}{\left(\frac{1}{\mathrm{x}}\right)}^{\mathrm{r}}$$$${\mathrm{T}}_{\mathrm{r}+1}=\left(\begin{array}{c}12\\ \mathrm{r}\end{array}\right){\left(\mathrm{x}\right)}^{12-\mathrm{r}-\mathrm{r}}$$$$\mathrm{As}{\mathrm{T}}_{\mathrm{r}+1}\mathrm{is}\mathrm{free}\mathrm{of}\mathrm{x}$$$$\mathrm{So}12-2\mathrm{r}=0\Rightarrow \mathrm{r}=6$$$${\mathrm{T}}_{\mathrm{r}+1}={\mathrm{T}}_{6+1}={\mathrm{T}}_7$$$${\mathrm{T}}_7\mathrm{is}\mathrm{free}\mathrm{of}\mathrm{x}$$

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