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3-Find-the-term-independ-ent-of-x-in-the-expansion-of-x-1-x-2-x-1-x-12-

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Question Number 27884 by das47955@mail.com last updated on 16/Jan/18
(3) Find the term independ− ent of x in the expansion of ( x+(1/x))^2 (x−(1/x))^(12)
$$\left(\mathrm{3}\right)\:\boldsymbol{\mathrm{Find}}\:\boldsymbol{\mathrm{the}}\:\boldsymbol{\mathrm{term}}\:\boldsymbol{\mathrm{independ}}− \\ $$$$\boldsymbol{\mathrm{ent}}\:\boldsymbol{\mathrm{of}}\:\:\boldsymbol{\mathrm{x}}\:\:\boldsymbol{\mathrm{in}}\:\boldsymbol{\mathrm{the}}\:\boldsymbol{\mathrm{expansion}}\:\boldsymbol{\mathrm{of}} \\ $$$$\:\:\:\left(\:\boldsymbol{\mathrm{x}}+\frac{\mathrm{1}}{\boldsymbol{\mathrm{x}}}\right)^{\mathrm{2}} \left(\boldsymbol{\mathrm{x}}−\frac{\mathrm{1}}{\boldsymbol{\mathrm{x}}}\right)^{\mathrm{12}} \\ $$
Answered by Rasheed.Sindhi last updated on 16/Jan/18
( x+(1/x))^2 (x−(1/x))^(12) (x^2 +2+(1/x^2 ))(x−(1/x))^(12) −−−− T_(r+1) = ((n),(r) ) a^(n−r) b^r −−−−−− T_(r+1) of (x−(1/x))^(12) T_(r+1) = (((12)),(( r)) ) (x)^(12−r) (−(1/x))^r T_(r+1) =(−1)^r (((12)),(( r)) ) (x)^(12−2r) T_(r+1) multiplied by (x^2 +2+(1/x^2 )) (x^2 +2+(1/x^2 )).T_(f+1) =x^2 T_(r+1) +2T_(r+1) +((1/x^2 ))T_(r+1) =x^2 (−1)^r (((12)),(( r)) ) (x)^(12−2r) +2(−1)^r (((12)),(( r)) ) (x)^(12−2r) +((1/x^2 ))(−1)^r (((12)),(( r)) ) (x)^(12−2r) =(−r)^r (((12)),(( r)) ) (x)^(14−2r) +2(−1)^r (((12)),(( r)) ) (x)^(12−2r) +(−1)^r (((12)),(( r)) ) (x)^(10−2r) When 14−2r=0⇒r=7 Free of x term 1st part of the above (−1)^7 (((12)),(( r)) ) (x)^(14−2(7)) =− (((12)),(( 7)) ) ..(i) When 12−2r=0⇒r=6 Free of x, 2nd part of above 2(−1)^6 (((12)),(( r)) ) (x)^(12−2r) =2 (((12)),(( 6)) )...(ii) When 10−2r=0⇒r=5 Free of x, 3rd part of the above (−1)^7 (((12)),(( r)) ) (x)^(10−2r) =− (((12)),(( 5)) ) ...(iii) Free of x term=(i)+(ii)+(iii) =− (((12)),(( 7)) ) +2 (((12)),(( 6)) ) − (((12)),(( 5)) ) =−792+2×924−792=264
$$\:\:\:\left(\:\boldsymbol{\mathrm{x}}+\frac{\mathrm{1}}{\boldsymbol{\mathrm{x}}}\right)^{\mathrm{2}} \left(\boldsymbol{\mathrm{x}}−\frac{\mathrm{1}}{\boldsymbol{\mathrm{x}}}\right)^{\mathrm{12}} \\ $$$$\:\:\left(\mathrm{x}^{\mathrm{2}} +\mathrm{2}+\frac{\mathrm{1}}{\mathrm{x}^{\mathrm{2}} }\right)\left(\boldsymbol{\mathrm{x}}−\frac{\mathrm{1}}{\boldsymbol{\mathrm{x}}}\right)^{\mathrm{12}} \\ $$$$−−−− \\ $$$$\mathrm{T}_{\mathrm{r}+\mathrm{1}} =\begin{pmatrix}{\mathrm{n}}\\{\mathrm{r}}\end{pmatrix}\:\mathrm{a}^{\mathrm{n}−\mathrm{r}} \mathrm{b}^{\mathrm{r}} \\ $$$$−−−−−− \\ $$$$\mathrm{T}_{\mathrm{r}+\mathrm{1}} \:\mathrm{of}\:\left(\boldsymbol{\mathrm{x}}−\frac{\mathrm{1}}{\boldsymbol{\mathrm{x}}}\right)^{\mathrm{12}} \\ $$$$\:\mathrm{T}_{\mathrm{r}+\mathrm{1}} =\begin{pmatrix}{\mathrm{12}}\\{\:\mathrm{r}}\end{pmatrix}\:\left(\mathrm{x}\right)^{\mathrm{12}−\mathrm{r}} \left(−\frac{\mathrm{1}}{\mathrm{x}}\right)^{\mathrm{r}} \\ $$$$\:\mathrm{T}_{\mathrm{r}+\mathrm{1}} =\left(−\mathrm{1}\right)^{\mathrm{r}} \begin{pmatrix}{\mathrm{12}}\\{\:\mathrm{r}}\end{pmatrix}\:\left(\mathrm{x}\right)^{\mathrm{12}−\mathrm{2r}} \\ $$$$\mathrm{T}_{\mathrm{r}+\mathrm{1}} \mathrm{multiplied}\:\mathrm{by}\:\left(\mathrm{x}^{\mathrm{2}} +\mathrm{2}+\frac{\mathrm{1}}{\mathrm{x}^{\mathrm{2}} }\right) \\ $$$$\left(\mathrm{x}^{\mathrm{2}} +\mathrm{2}+\frac{\mathrm{1}}{\mathrm{x}^{\mathrm{2}} }\right).\mathrm{T}_{\mathrm{f}+\mathrm{1}} =\mathrm{x}^{\mathrm{2}} \mathrm{T}_{\mathrm{r}+\mathrm{1}} +\mathrm{2T}_{\mathrm{r}+\mathrm{1}} +\left(\frac{\mathrm{1}}{\mathrm{x}^{\mathrm{2}} }\right)\mathrm{T}_{\mathrm{r}+\mathrm{1}} \\ $$$$\:\:\:\:\:=\mathrm{x}^{\mathrm{2}} \left(−\mathrm{1}\right)^{\mathrm{r}} \begin{pmatrix}{\mathrm{12}}\\{\:\mathrm{r}}\end{pmatrix}\:\left(\mathrm{x}\right)^{\mathrm{12}−\mathrm{2r}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:+\mathrm{2}\left(−\mathrm{1}\right)^{\mathrm{r}} \begin{pmatrix}{\mathrm{12}}\\{\:\mathrm{r}}\end{pmatrix}\:\left(\mathrm{x}\right)^{\mathrm{12}−\mathrm{2r}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:+\left(\frac{\mathrm{1}}{\mathrm{x}^{\mathrm{2}} }\right)\left(−\mathrm{1}\right)^{\mathrm{r}} \begin{pmatrix}{\mathrm{12}}\\{\:\mathrm{r}}\end{pmatrix}\:\left(\mathrm{x}\right)^{\mathrm{12}−\mathrm{2r}} \\ $$$$\:\:\:=\left(−\mathrm{r}\right)^{\mathrm{r}} \begin{pmatrix}{\mathrm{12}}\\{\:\mathrm{r}}\end{pmatrix}\:\left(\mathrm{x}\right)^{\mathrm{14}−\mathrm{2r}} \\ $$$$\:\:\:\:\:\:\:\:+\mathrm{2}\left(−\mathrm{1}\right)^{\mathrm{r}} \begin{pmatrix}{\mathrm{12}}\\{\:\mathrm{r}}\end{pmatrix}\:\left(\mathrm{x}\right)^{\mathrm{12}−\mathrm{2r}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:+\left(−\mathrm{1}\right)^{\mathrm{r}} \begin{pmatrix}{\mathrm{12}}\\{\:\mathrm{r}}\end{pmatrix}\:\left(\mathrm{x}\right)^{\mathrm{10}−\mathrm{2r}} \\ $$$$\mathrm{When}\:\mathrm{14}−\mathrm{2r}=\mathrm{0}\Rightarrow\mathrm{r}=\mathrm{7} \\ $$$$\mathrm{Free}\:\mathrm{of}\:\mathrm{x}\:\mathrm{term}\:\mathrm{1st}\:\mathrm{part}\:\mathrm{of}\:\mathrm{the}\:\mathrm{above} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\left(−\mathrm{1}\right)^{\mathrm{7}} \begin{pmatrix}{\mathrm{12}}\\{\:\mathrm{r}}\end{pmatrix}\:\left(\mathrm{x}\right)^{\mathrm{14}−\mathrm{2}\left(\mathrm{7}\right)} =−\begin{pmatrix}{\mathrm{12}}\\{\:\mathrm{7}}\end{pmatrix}\:..\left(\mathrm{i}\right) \\ $$$$\mathrm{When}\:\mathrm{12}−\mathrm{2r}=\mathrm{0}\Rightarrow\mathrm{r}=\mathrm{6} \\ $$$$\mathrm{Free}\:\mathrm{of}\:\mathrm{x},\:\mathrm{2nd}\:\mathrm{part}\:\mathrm{of}\:\mathrm{above}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\: \\ $$$$\:\:\:\:\:\:\mathrm{2}\left(−\mathrm{1}\right)^{\mathrm{6}} \begin{pmatrix}{\mathrm{12}}\\{\:\mathrm{r}}\end{pmatrix}\:\left(\mathrm{x}\right)^{\mathrm{12}−\mathrm{2r}} =\mathrm{2}\begin{pmatrix}{\mathrm{12}}\\{\:\mathrm{6}}\end{pmatrix}…\left(\mathrm{ii}\right) \\ $$$$\mathrm{When}\:\mathrm{10}−\mathrm{2r}=\mathrm{0}\Rightarrow\mathrm{r}=\mathrm{5} \\ $$$$\mathrm{Free}\:\mathrm{of}\:\mathrm{x},\:\mathrm{3rd}\:\mathrm{part}\:\mathrm{of}\:\mathrm{the}\:\mathrm{above} \\ $$$$\:\:\:\:\:\:\:\left(−\mathrm{1}\right)^{\mathrm{7}} \begin{pmatrix}{\mathrm{12}}\\{\:\mathrm{r}}\end{pmatrix}\:\left(\mathrm{x}\right)^{\mathrm{10}−\mathrm{2r}} =−\begin{pmatrix}{\mathrm{12}}\\{\:\mathrm{5}}\end{pmatrix}\:…\left(\mathrm{iii}\right) \\ $$$$\mathrm{Free}\:\mathrm{of}\:\mathrm{x}\:\mathrm{term}=\left(\mathrm{i}\right)+\left(\mathrm{ii}\right)+\left(\mathrm{iii}\right) \\ $$$$\:\:\:\:\:\:\:=−\begin{pmatrix}{\mathrm{12}}\\{\:\mathrm{7}}\end{pmatrix}\:+\mathrm{2}\begin{pmatrix}{\mathrm{12}}\\{\:\mathrm{6}}\end{pmatrix}\:−\begin{pmatrix}{\mathrm{12}}\\{\:\mathrm{5}}\end{pmatrix}\: \\ $$$$ \\ $$$$\:\:\:\:\:\:\:\:=−\mathrm{792}+\mathrm{2}×\mathrm{924}−\mathrm{792}=\mathrm{264} \\ $$
Commented by Rasheed.Sindhi last updated on 16/Jan/18
Corrected now.
$$\mathrm{Corrected}\:\mathrm{now}. \\ $$
Answered by mrW2 last updated on 16/Jan/18
( x+(1/x))^2 (x−(1/x))^(12) =(1/x^(14) )( x^2 +1)^2 (x^2 −1)^(12) =(1/x^(14) )( x^4 +2x^2 +1)(x^2 −1)^(12) =(1/x^(14) )( x^4 +2x^2 +1)Σ_(k=0) ^(12) C_k ^(12) x^(2k) (−1)^(12−k) term independent of x is: (1/x^(14) )×x^4 ×C_5 ^(12) x^(2×5) (−1)^(12−5) +(1/x^(14) )×2x^2 ×C_6 ^(12) x^(2×6) (−1)^(12−6) +(1/x^(14) )×1×C_7 ^(12) x^(2×7) (−1)^(12−7) =−C_5 ^(12) +2C_6 ^(12) −C_7 ^(12) =−792+2×924−792 =264
$$\left(\:\boldsymbol{\mathrm{x}}+\frac{\mathrm{1}}{\boldsymbol{\mathrm{x}}}\right)^{\mathrm{2}} \left(\boldsymbol{\mathrm{x}}−\frac{\mathrm{1}}{\boldsymbol{\mathrm{x}}}\right)^{\mathrm{12}} \\ $$$$=\frac{\mathrm{1}}{{x}^{\mathrm{14}} }\left(\:\boldsymbol{\mathrm{x}}^{\mathrm{2}} +\mathrm{1}\right)^{\mathrm{2}} \left(\boldsymbol{\mathrm{x}}^{\mathrm{2}} −\mathrm{1}\right)^{\mathrm{12}} \\ $$$$=\frac{\mathrm{1}}{{x}^{\mathrm{14}} }\left(\:\boldsymbol{\mathrm{x}}^{\mathrm{4}} +\mathrm{2}{x}^{\mathrm{2}} +\mathrm{1}\right)\left(\boldsymbol{\mathrm{x}}^{\mathrm{2}} −\mathrm{1}\right)^{\mathrm{12}} \\ $$$$=\frac{\mathrm{1}}{{x}^{\mathrm{14}} }\left(\:\boldsymbol{\mathrm{x}}^{\mathrm{4}} +\mathrm{2}{x}^{\mathrm{2}} +\mathrm{1}\right)\underset{{k}=\mathrm{0}} {\overset{\mathrm{12}} {\sum}}{C}_{{k}} ^{\mathrm{12}} {x}^{\mathrm{2}{k}} \left(−\mathrm{1}\right)^{\mathrm{12}−{k}} \\ $$$${term}\:{independent}\:{of}\:{x}\:{is}: \\ $$$$\frac{\mathrm{1}}{{x}^{\mathrm{14}} }×{x}^{\mathrm{4}} ×{C}_{\mathrm{5}} ^{\mathrm{12}} {x}^{\mathrm{2}×\mathrm{5}} \left(−\mathrm{1}\right)^{\mathrm{12}−\mathrm{5}} +\frac{\mathrm{1}}{{x}^{\mathrm{14}} }×\mathrm{2}{x}^{\mathrm{2}} ×{C}_{\mathrm{6}} ^{\mathrm{12}} {x}^{\mathrm{2}×\mathrm{6}} \left(−\mathrm{1}\right)^{\mathrm{12}−\mathrm{6}} +\frac{\mathrm{1}}{{x}^{\mathrm{14}} }×\mathrm{1}×{C}_{\mathrm{7}} ^{\mathrm{12}} {x}^{\mathrm{2}×\mathrm{7}} \left(−\mathrm{1}\right)^{\mathrm{12}−\mathrm{7}} \\ $$$$=−{C}_{\mathrm{5}} ^{\mathrm{12}} +\mathrm{2}{C}_{\mathrm{6}} ^{\mathrm{12}} −{C}_{\mathrm{7}} ^{\mathrm{12}} \\ $$$$=−\mathrm{792}+\mathrm{2}×\mathrm{924}−\mathrm{792} \\ $$$$=\mathrm{264} \\ $$
Commented by mrW2 last updated on 16/Jan/18

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