Question and Answers Forum

All Questions      Topic List

Others Questions

Previous in All Question      Next in All Question      

Previous in Others      Next in Others      

Question Number 27809 by das47955@mail.com last updated on 15/Jan/18

(1) Find the term independent  of   x   in the  expansion of                         (x−(2/x))^(10)

$$\left(\mathrm{1}\right)\:\boldsymbol{\mathrm{Find}}\:\boldsymbol{\mathrm{the}}\:\boldsymbol{\mathrm{term}}\:\boldsymbol{\mathrm{independent}} \\ $$$$\boldsymbol{\mathrm{of}}\:\:\:\boldsymbol{\mathrm{x}}\:\:\:\boldsymbol{\mathrm{in}}\:\boldsymbol{\mathrm{the}}\:\:\boldsymbol{\mathrm{expansion}}\:\boldsymbol{\mathrm{of}}\: \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\left(\boldsymbol{\mathrm{x}}−\frac{\mathrm{2}}{\boldsymbol{\mathrm{x}}}\right)^{\mathrm{10}} \\ $$

Answered by 8/mln(naing)060691 last updated on 15/Jan/18

(r+1)^(th) term=^(10) C_r x^(10−r) (−(2/x))^r                             =^(10) C_r .x^(10−r) (−2)^r x^(−r)                             =^(10) C_r (−2)^r x^(10−2r)   To get the term independent of x,  put 10−2r=0⇒r=5  ∴the term independent of x=^(10) C_5 (−2)^5

$$\left(\mathrm{r}+\mathrm{1}\right)^{\mathrm{th}} \mathrm{term}=^{\mathrm{10}} \mathrm{C}_{\mathrm{r}} \mathrm{x}^{\mathrm{10}−\mathrm{r}} \left(−\frac{\mathrm{2}}{\mathrm{x}}\right)^{\mathrm{r}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=^{\mathrm{10}} \mathrm{C}_{\mathrm{r}} .\mathrm{x}^{\mathrm{10}−\mathrm{r}} \left(−\mathrm{2}\right)^{\mathrm{r}} \mathrm{x}^{−\mathrm{r}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=^{\mathrm{10}} \mathrm{C}_{\mathrm{r}} \left(−\mathrm{2}\right)^{\mathrm{r}} \mathrm{x}^{\mathrm{10}−\mathrm{2r}} \\ $$$$\mathrm{To}\:\mathrm{get}\:\mathrm{the}\:\mathrm{term}\:\mathrm{independent}\:\mathrm{of}\:\mathrm{x}, \\ $$$$\mathrm{put}\:\mathrm{10}−\mathrm{2r}=\mathrm{0}\Rightarrow\mathrm{r}=\mathrm{5} \\ $$$$\therefore\mathrm{the}\:\mathrm{term}\:\mathrm{independent}\:\mathrm{of}\:\mathrm{x}=^{\mathrm{10}} \mathrm{C}_{\mathrm{5}} \left(−\mathrm{2}\right)^{\mathrm{5}} \\ $$

Commented by das47955@mail.com last updated on 15/Jan/18

  wow.i need this process

$$ \\ $$$$\mathrm{wow}.\mathrm{i}\:\mathrm{need}\:\mathrm{this}\:\mathrm{process} \\ $$

Commented by das47955@mail.com last updated on 15/Jan/18

  really thanks .....

$$ \\ $$$$\mathrm{really}\:\mathrm{thanks}\:..... \\ $$

Answered by mrW2 last updated on 15/Jan/18

(x−(2/x))^(10) =Σ_(k=0) ^(10) C_k ^(10) x^k (−(2/x))^(10−k)   =Σ_(k=0) ^(10) C_k ^(10) (−2)^(10−k) x^(2k−10)   2k−10=0  k=5  x−independent term is  C_5 ^(10) (−2)^5 =−8064

$$\left({x}−\frac{\mathrm{2}}{{x}}\right)^{\mathrm{10}} =\underset{{k}=\mathrm{0}} {\overset{\mathrm{10}} {\sum}}{C}_{{k}} ^{\mathrm{10}} {x}^{{k}} \left(−\frac{\mathrm{2}}{{x}}\right)^{\mathrm{10}−{k}} \\ $$$$=\underset{{k}=\mathrm{0}} {\overset{\mathrm{10}} {\sum}}{C}_{{k}} ^{\mathrm{10}} \left(−\mathrm{2}\right)^{\mathrm{10}−{k}} {x}^{\mathrm{2}{k}−\mathrm{10}} \\ $$$$\mathrm{2}{k}−\mathrm{10}=\mathrm{0} \\ $$$${k}=\mathrm{5} \\ $$$${x}−{independent}\:{term}\:{is} \\ $$$${C}_{\mathrm{5}} ^{\mathrm{10}} \left(−\mathrm{2}\right)^{\mathrm{5}} =−\mathrm{8064} \\ $$

Commented by das47955@mail.com last updated on 15/Jan/18

Thanks ....

$$\boldsymbol{\mathrm{T}}\mathrm{hanks}\:.... \\ $$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com