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

(2) Find the term indepen− dent of x in the expansion of (2x^2 +(1/x))^6

$$\left(\mathrm{2}\right)\:\:\boldsymbol{\mathrm{Find}}\:\boldsymbol{\mathrm{the}}\:\boldsymbol{\mathrm{term}}\:\boldsymbol{\mathrm{indepen}}− \\ $$$$\boldsymbol{\mathrm{dent}}\:\boldsymbol{\mathrm{of}}\:\:\boldsymbol{\mathrm{x}}\:\:\boldsymbol{\mathrm{in}}\:\boldsymbol{\mathrm{the}}\:\boldsymbol{\mathrm{expansion}}\:\boldsymbol{\mathrm{of}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\left(\mathrm{2}\boldsymbol{\mathrm{x}}^{\mathrm{2}} +\frac{\mathrm{1}}{\boldsymbol{\mathrm{x}}}\right)^{\mathrm{6}} \\ $$

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

T_(r+1) = ((n),(r) ) a^(n−r) b^r T_(r+1) = ((6),(r) ) (2x^2 )^(6−r) ((1/x))^r = ((6),(r) ) (2)^(6−r) (x^(12−2r) )((1/x^r )) = ((6),(r) ) (2)^(6−r) (x^(12−3r) ) If T_(r+1) is free of x 12−3r=0⇒r=4 T_(r+1) =T_(4+1) =T_5 T_5 is free of x

$$\mathrm{T}_{\mathrm{r}+\mathrm{1}} =\begin{pmatrix}{\mathrm{n}}\\{\mathrm{r}}\end{pmatrix}\:\mathrm{a}^{\mathrm{n}−\mathrm{r}} \mathrm{b}^{\mathrm{r}} \\ $$$$\mathrm{T}_{\mathrm{r}+\mathrm{1}} =\begin{pmatrix}{\mathrm{6}}\\{\mathrm{r}}\end{pmatrix}\:\left(\mathrm{2x}^{\mathrm{2}} \right)^{\mathrm{6}−\mathrm{r}} \left(\frac{\mathrm{1}}{\mathrm{x}}\right)^{\mathrm{r}} \\ $$$$\:\:\:\:\:\:\:\:=\begin{pmatrix}{\mathrm{6}}\\{\mathrm{r}}\end{pmatrix}\:\left(\mathrm{2}\right)^{\mathrm{6}−\mathrm{r}} \left(\mathrm{x}^{\mathrm{12}−\mathrm{2r}} \right)\left(\frac{\mathrm{1}}{\mathrm{x}^{\mathrm{r}} }\right) \\ $$$$\:\:\:\:\:\:\:\:=\begin{pmatrix}{\mathrm{6}}\\{\mathrm{r}}\end{pmatrix}\:\left(\mathrm{2}\right)^{\mathrm{6}−\mathrm{r}} \left(\mathrm{x}^{\mathrm{12}−\mathrm{3r}} \right) \\ $$$$\mathrm{If}\:\mathrm{T}_{\mathrm{r}+\mathrm{1}} \mathrm{is}\:\mathrm{free}\:\mathrm{of}\:\mathrm{x} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{12}−\mathrm{3r}=\mathrm{0}\Rightarrow\mathrm{r}=\mathrm{4} \\ $$$$\:\:\:\:\:\:\mathrm{T}_{\mathrm{r}+\mathrm{1}} =\mathrm{T}_{\mathrm{4}+\mathrm{1}} =\mathrm{T}_{\mathrm{5}} \\ $$$$\mathrm{T}_{\mathrm{5}} \:\mathrm{is}\:\mathrm{free}\:\mathrm{of}\:\mathrm{x} \\ $$

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