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4-Find-the-term-indepen-dent-of-x-in-the-expansion-of-x-2-2-1-x-2-6-




Question Number 27885 by das47955@mail.com last updated on 16/Jan/18
(4)  Find the term indepen−  dent of    x   in the expansion of            (x^2 −2+(1/x^2 ))^6
$$\left(\mathrm{4}\right)\:\:\boldsymbol{\mathrm{Find}}\:\boldsymbol{\mathrm{the}}\:\boldsymbol{\mathrm{term}}\:\boldsymbol{\mathrm{indepen}}− \\ $$$$\boldsymbol{\mathrm{dent}}\:\boldsymbol{\mathrm{of}}\:\:\:\:\boldsymbol{\mathrm{x}}\:\:\:\boldsymbol{\mathrm{in}}\:\boldsymbol{\mathrm{the}}\:\boldsymbol{\mathrm{expansion}}\:\boldsymbol{\mathrm{of}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\left(\boldsymbol{\mathrm{x}}^{\mathrm{2}} −\mathrm{2}+\frac{\mathrm{1}}{\boldsymbol{\mathrm{x}}^{\mathrm{2}} }\right)^{\mathrm{6}} \\ $$
Commented by Rasheed.Sindhi last updated on 16/Jan/18
x^2 −2+(1/x^2 )=(x−(1/x))^2   (x^2 −2+(1/x^2 ))^6 ={(x−(1/x))^2 }^6   =(x−(1/x))^(12)   T_(r+1) = ((n),(r) ) a^(n−r) b^r    T_(r+1) = (((12)),(r) ) (x)^(12−r) ((1/x))^r    T_(r+1) = (((12)),(r) ) (x)^(12−r−r)   As T_(r+1)  is free of  x  So  12−2r=0⇒r=6  T_(r+1) =T_(6+1) =T_7   T_7  is free of x
$$\boldsymbol{\mathrm{x}}^{\mathrm{2}} −\mathrm{2}+\frac{\mathrm{1}}{\boldsymbol{\mathrm{x}}^{\mathrm{2}} }=\left(\mathrm{x}−\frac{\mathrm{1}}{\mathrm{x}}\right)^{\mathrm{2}} \\ $$$$\left(\boldsymbol{\mathrm{x}}^{\mathrm{2}} −\mathrm{2}+\frac{\mathrm{1}}{\boldsymbol{\mathrm{x}}^{\mathrm{2}} }\right)^{\mathrm{6}} =\left\{\left(\mathrm{x}−\frac{\mathrm{1}}{\mathrm{x}}\right)^{\mathrm{2}} \right\}^{\mathrm{6}} \\ $$$$=\left(\mathrm{x}−\frac{\mathrm{1}}{\mathrm{x}}\right)^{\mathrm{12}} \\ $$$$\mathrm{T}_{\mathrm{r}+\mathrm{1}} =\begin{pmatrix}{\mathrm{n}}\\{\mathrm{r}}\end{pmatrix}\:\mathrm{a}^{\mathrm{n}−\mathrm{r}} \mathrm{b}^{\mathrm{r}} \\ $$$$\:\mathrm{T}_{\mathrm{r}+\mathrm{1}} =\begin{pmatrix}{\mathrm{12}}\\{\mathrm{r}}\end{pmatrix}\:\left(\mathrm{x}\right)^{\mathrm{12}−\mathrm{r}} \left(\frac{\mathrm{1}}{\mathrm{x}}\right)^{\mathrm{r}} \\ $$$$\:\mathrm{T}_{\mathrm{r}+\mathrm{1}} =\begin{pmatrix}{\mathrm{12}}\\{\mathrm{r}}\end{pmatrix}\:\left(\mathrm{x}\right)^{\mathrm{12}−\mathrm{r}−\mathrm{r}} \\ $$$$\mathrm{As}\:\mathrm{T}_{\mathrm{r}+\mathrm{1}} \:\mathrm{is}\:\mathrm{free}\:\mathrm{of}\:\:\mathrm{x} \\ $$$$\mathrm{So}\:\:\mathrm{12}−\mathrm{2r}=\mathrm{0}\Rightarrow\mathrm{r}=\mathrm{6} \\ $$$$\mathrm{T}_{\mathrm{r}+\mathrm{1}} =\mathrm{T}_{\mathrm{6}+\mathrm{1}} =\mathrm{T}_{\mathrm{7}} \\ $$$$\mathrm{T}_{\mathrm{7}} \:\mathrm{is}\:\mathrm{free}\:\mathrm{of}\:\mathrm{x} \\ $$

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