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Question Number 131248 by bramlexs22 last updated on 03/Feb/21
![If α and β are the coefficient of x^8 and x^(−24) respectively in the expansion of [ x^4 +2+(1/x^4 ) ]^(10) in powers of x then (α/β) is equal to](Q131248.png)
$${If}\:\alpha\:{and}\:\beta\:{are}\:{the}\:{coefficient}\: \\ $$$${of}\:{x}^{\mathrm{8}} \:{and}\:{x}^{−\mathrm{24}} \:{respectively}\: \\ $$$${in}\:{the}\:{expansion}\:{of}\:\left[\:{x}^{\mathrm{4}} +\mathrm{2}+\frac{\mathrm{1}}{{x}^{\mathrm{4}} }\:\right]^{\mathrm{10}} \\ $$$${in}\:{powers}\:{of}\:{x}\:{then}\:\frac{\alpha}{\beta}\:{is}\:{equal}\:{to}\: \\ $$
Answered by EDWIN88 last updated on 03/Feb/21
![[(x^4 +x^(−4) )+2 ]^(10) = Σ_(k=0) ^(20) (((20)),(( k)) ) (x^4 +x^(−4) )^(20−k) (2)^k = Σ_(k=0) ^(20) (((20)),(( k)) ) 2^k [ Σ_(k=0) ^(20−k) (((20−k)),(( k)) )(x^4 )^(20−k) (x^(−4) )^k ]](Q131258.png)
$$\:\left[\left({x}^{\mathrm{4}} +{x}^{−\mathrm{4}} \right)+\mathrm{2}\:\right]^{\mathrm{10}} =\:\underset{{k}=\mathrm{0}} {\overset{\mathrm{20}} {\sum}}\:\begin{pmatrix}{\mathrm{20}}\\{\:\:{k}}\end{pmatrix}\:\left({x}^{\mathrm{4}} +{x}^{−\mathrm{4}} \right)^{\mathrm{20}−{k}} \:\left(\mathrm{2}\right)^{{k}} \\ $$$$\:=\:\underset{{k}=\mathrm{0}} {\overset{\mathrm{20}} {\sum}}\:\begin{pmatrix}{\mathrm{20}}\\{\:\:{k}}\end{pmatrix}\:\mathrm{2}^{{k}} \:\left[\:\underset{{k}=\mathrm{0}} {\overset{\mathrm{20}−{k}} {\sum}}\:\begin{pmatrix}{\mathrm{20}−{k}}\\{\:\:\:\:\:{k}}\end{pmatrix}\left({x}^{\mathrm{4}} \right)^{\mathrm{20}−{k}} \:\left({x}^{−\mathrm{4}} \right)^{{k}} \:\right] \\ $$
Answered by mr W last updated on 03/Feb/21
![[x^4 +2+(1/x^4 )]^(10) =(1/x^(40) )[x^8 +2x^4 +1]^(10) =(1/x^(40) )(1+x^4 )^(20) =(1/x^(40) )Σ_(k=0) ^(20) C_k ^(20) x^(4k) =Σ_(k=0) ^(20) C_k ^(20) x^(4k−40) term x^8 : 4k−40=8 ⇒k=12 ⇒α=C_(12) ^(20) term x^(−24) : 4k−40=−24 ⇒k=4 ⇒β=C_4 ^(20) (α/β)=(C_(12) ^(20) /C_4 ^(20) )=((20!)/(12!8!))×((4!×16!)/(20!))=((16×15×14×13)/(8×7×6×5)) =26](Q131268.png)
$$\left[{x}^{\mathrm{4}} +\mathrm{2}+\frac{\mathrm{1}}{{x}^{\mathrm{4}} }\right]^{\mathrm{10}} \\ $$$$=\frac{\mathrm{1}}{{x}^{\mathrm{40}} }\left[{x}^{\mathrm{8}} +\mathrm{2}{x}^{\mathrm{4}} +\mathrm{1}\right]^{\mathrm{10}} \\ $$$$=\frac{\mathrm{1}}{{x}^{\mathrm{40}} }\left(\mathrm{1}+{x}^{\mathrm{4}} \right)^{\mathrm{20}} \\ $$$$=\frac{\mathrm{1}}{{x}^{\mathrm{40}} }\underset{{k}=\mathrm{0}} {\overset{\mathrm{20}} {\sum}}{C}_{{k}} ^{\mathrm{20}} {x}^{\mathrm{4}{k}} \\ $$$$=\underset{{k}=\mathrm{0}} {\overset{\mathrm{20}} {\sum}}{C}_{{k}} ^{\mathrm{20}} {x}^{\mathrm{4}{k}−\mathrm{40}} \\ $$$${term}\:{x}^{\mathrm{8}} :\:\mathrm{4}{k}−\mathrm{40}=\mathrm{8}\:\Rightarrow{k}=\mathrm{12}\:\Rightarrow\alpha={C}_{\mathrm{12}} ^{\mathrm{20}} \\ $$$${term}\:{x}^{−\mathrm{24}} :\:\mathrm{4}{k}−\mathrm{40}=−\mathrm{24}\:\Rightarrow{k}=\mathrm{4}\:\Rightarrow\beta={C}_{\mathrm{4}} ^{\mathrm{20}} \\ $$$$\frac{\alpha}{\beta}=\frac{{C}_{\mathrm{12}} ^{\mathrm{20}} }{{C}_{\mathrm{4}} ^{\mathrm{20}} }=\frac{\mathrm{20}!}{\mathrm{12}!\mathrm{8}!}×\frac{\mathrm{4}!×\mathrm{16}!}{\mathrm{20}!}=\frac{\mathrm{16}×\mathrm{15}×\mathrm{14}×\mathrm{13}}{\mathrm{8}×\mathrm{7}×\mathrm{6}×\mathrm{5}} \\ $$$$=\mathrm{26} \\ $$
Commented by EDWIN88 last updated on 03/Feb/21
$${nice} \\ $$