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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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