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Question Number 131688 by liberty last updated on 07/Feb/21

Let α and β are the roots of the equation x^2 −6x−2=0. If a_n = α^n −β^n for n ≥1 then the value of ((a_(10) −2a_8 )/(2a_9 )) ?

$$\:\mathrm{Let}\:\alpha\:\mathrm{and}\:\beta\:\mathrm{are}\:\mathrm{the}\:\mathrm{roots}\: \\ $$$$\mathrm{of}\:\mathrm{the}\:\mathrm{equation}\:\mathrm{x}^{\mathrm{2}} −\mathrm{6x}−\mathrm{2}=\mathrm{0}. \\ $$$$\mathrm{If}\:{a}_{{n}} \:=\:\alpha^{{n}} −\beta^{{n}} \:\mathrm{for}\:{n}\:\geqslant\mathrm{1}\: \\ $$$$\mathrm{then}\:\mathrm{the}\:\mathrm{value}\:\mathrm{of}\:\frac{{a}_{\mathrm{10}} −\mathrm{2}{a}_{\mathrm{8}} }{\mathrm{2}{a}_{\mathrm{9}} }\:? \\ $$

Answered by mathmax by abdo last updated on 07/Feb/21

x^2 −6x−2=0 →Δ^′ =9+2=11 ⇒x_1 =3+(√(11)) and x_2 =3−(√(11)) a_n =x_1 ^n −x_2 ^n =(3+(√(11)))^n −(3−(√(11)))^n =Σ_(k=0) ^n C_n ^k 3^(n−k) ((√(11)))^k −Σ_(k=0) ^n C_n ^k 3^(n−k) (−(√(11)))^k =Σ_(k=0) ^n C_n ^k 3^(n−k) { 1−(−1)^k )((√(11)))^k =Σ_(p=0) ^([((n−1)/2)]) C_n ^(2p+1) 3^(n−2p−1) 2.((√(11)))^(2p+1) =2(√(11))Σ_(p=0) ^([((n−1)/2)]) C_n ^(2p+1) 3^(n−2p−1) (11)^p ⇒a_(10) =2(√(11))Σ_(p=0) ^4 C_(10) ^(2p+1) 3^(9−2p) (11)^p a_8 =2(√(11))Σ_(p=0) ^3 C_8 ^(2p+1) 3^(7−2p) (11)^p a_9 =2(√(11))Σ_(p=0) ^4 C_9 ^(2p+1) 3^(8−2p) (11)^p rest to finish the calculus...

$$\mathrm{x}^{\mathrm{2}} −\mathrm{6x}−\mathrm{2}=\mathrm{0}\:\rightarrow\Delta^{'} \:=\mathrm{9}+\mathrm{2}=\mathrm{11}\:\Rightarrow\mathrm{x}_{\mathrm{1}} =\mathrm{3}+\sqrt{\mathrm{11}}\:\mathrm{and}\:\mathrm{x}_{\mathrm{2}} =\mathrm{3}−\sqrt{\mathrm{11}} \\ $$$$\mathrm{a}_{\mathrm{n}} =\mathrm{x}_{\mathrm{1}} ^{\mathrm{n}} −\mathrm{x}_{\mathrm{2}} ^{\mathrm{n}} \:=\left(\mathrm{3}+\sqrt{\mathrm{11}}\right)^{\mathrm{n}} −\left(\mathrm{3}−\sqrt{\mathrm{11}}\right)^{\mathrm{n}} \\ $$$$=\sum_{\mathrm{k}=\mathrm{0}} ^{\mathrm{n}} \:\mathrm{C}_{\mathrm{n}} ^{\mathrm{k}} \:\mathrm{3}^{\mathrm{n}−\mathrm{k}} \left(\sqrt{\mathrm{11}}\right)^{\mathrm{k}} −\sum_{\mathrm{k}=\mathrm{0}} ^{\mathrm{n}} \mathrm{C}_{\mathrm{n}} ^{\mathrm{k}} \:\mathrm{3}^{\mathrm{n}−\mathrm{k}} \left(−\sqrt{\mathrm{11}}\right)^{\mathrm{k}} \\ $$$$=\sum_{\mathrm{k}=\mathrm{0}} ^{\mathrm{n}} \:\mathrm{C}_{\mathrm{n}} ^{\mathrm{k}} \:\mathrm{3}^{\mathrm{n}−\mathrm{k}} \left\{\:\mathrm{1}−\left(−\mathrm{1}\right)^{\mathrm{k}} \right)\left(\sqrt{\mathrm{11}}\right)^{\mathrm{k}} \\ $$$$=\sum_{\mathrm{p}=\mathrm{0}} ^{\left[\frac{\mathrm{n}−\mathrm{1}}{\mathrm{2}}\right]} \:\mathrm{C}_{\mathrm{n}} ^{\mathrm{2p}+\mathrm{1}} \:\mathrm{3}^{\mathrm{n}−\mathrm{2p}−\mathrm{1}} \mathrm{2}.\left(\sqrt{\mathrm{11}}\right)^{\mathrm{2p}+\mathrm{1}} \\ $$$$=\mathrm{2}\sqrt{\mathrm{11}}\sum_{\mathrm{p}=\mathrm{0}} ^{\left[\frac{\mathrm{n}−\mathrm{1}}{\mathrm{2}}\right]} \:\mathrm{C}_{\mathrm{n}} ^{\mathrm{2p}+\mathrm{1}} \:\mathrm{3}^{\mathrm{n}−\mathrm{2p}−\mathrm{1}} \:\left(\mathrm{11}\right)^{\mathrm{p}} \\ $$$$\Rightarrow\mathrm{a}_{\mathrm{10}} =\mathrm{2}\sqrt{\mathrm{11}}\sum_{\mathrm{p}=\mathrm{0}} ^{\mathrm{4}} \:\mathrm{C}_{\mathrm{10}} ^{\mathrm{2p}+\mathrm{1}} \:\mathrm{3}^{\mathrm{9}−\mathrm{2p}} \left(\mathrm{11}\right)^{\mathrm{p}} \\ $$$$\mathrm{a}_{\mathrm{8}} =\mathrm{2}\sqrt{\mathrm{11}}\sum_{\mathrm{p}=\mathrm{0}} ^{\mathrm{3}} \mathrm{C}_{\mathrm{8}} ^{\mathrm{2p}+\mathrm{1}} \:\mathrm{3}^{\mathrm{7}−\mathrm{2p}} \:\left(\mathrm{11}\right)^{\mathrm{p}} \\ $$$$\mathrm{a}_{\mathrm{9}} =\mathrm{2}\sqrt{\mathrm{11}}\sum_{\mathrm{p}=\mathrm{0}} ^{\mathrm{4}} \:\mathrm{C}_{\mathrm{9}} ^{\mathrm{2p}+\mathrm{1}} \:\mathrm{3}^{\mathrm{8}−\mathrm{2p}} \left(\mathrm{11}\right)^{\mathrm{p}} \:\mathrm{rest}\:\mathrm{to}\:\mathrm{finish}\:\mathrm{the}\:\mathrm{calculus}... \\ $$

Answered by bemath last updated on 07/Feb/21

x^2 −6x−2=0→ { (α),(β) :} ⇒α^2 −6α−2=0...(×α^8 ) ⇒α^(10) −6α^9 −2α^8 =0⇒2α^8 =α^(10) −6α^9 ⇒β^(10) −6β^9 −2β^9 =0⇒2β^8 =β^(10) −6β^9 (1)−(2)⇒2(α^8 −β^8 )=α^(10) −β^(10) −6α^9 +6β^9 a_(10) = α^(10) −β^(10) ; a_8 =α^8 −β^8 ; a_9 = α^9 −β^9 we find ((a_(10) −2a_8 )/(2a_9 )) = ((α^(10) −β^(10) −2(α^8 −β^8 ))/(2(α^9 −β^9 ))) = ((α^(10) −β^(10) −(α^(10) −β^(10) −6α^9 +6β^9 ))/(2(α^9 −β^9 )))= 3

$$\:\mathrm{x}^{\mathrm{2}} −\mathrm{6x}−\mathrm{2}=\mathrm{0}\rightarrow\begin{cases}{\alpha}\\{\beta}\end{cases} \\ $$$$\:\Rightarrow\alpha^{\mathrm{2}} −\mathrm{6}\alpha−\mathrm{2}=\mathrm{0}...\left(×\alpha^{\mathrm{8}} \right) \\ $$$$\Rightarrow\alpha^{\mathrm{10}} −\mathrm{6}\alpha^{\mathrm{9}} −\mathrm{2}\alpha^{\mathrm{8}} =\mathrm{0}\Rightarrow\mathrm{2}\alpha^{\mathrm{8}} =\alpha^{\mathrm{10}} −\mathrm{6}\alpha^{\mathrm{9}} \\ $$$$\Rightarrow\beta^{\mathrm{10}} −\mathrm{6}\beta^{\mathrm{9}} −\mathrm{2}\beta^{\mathrm{9}} =\mathrm{0}\Rightarrow\mathrm{2}\beta^{\mathrm{8}} =\beta^{\mathrm{10}} −\mathrm{6}\beta^{\mathrm{9}} \\ $$$$\left(\mathrm{1}\right)−\left(\mathrm{2}\right)\Rightarrow\mathrm{2}\left(\alpha^{\mathrm{8}} −\beta^{\mathrm{8}} \right)=\alpha^{\mathrm{10}} −\beta^{\mathrm{10}} −\mathrm{6}\alpha^{\mathrm{9}} +\mathrm{6}\beta^{\mathrm{9}} \\ $$$${a}_{\mathrm{10}} =\:\alpha^{\mathrm{10}} −\beta^{\mathrm{10}} \:;\:{a}_{\mathrm{8}} =\alpha^{\mathrm{8}} −\beta^{\mathrm{8}} \:;\:{a}_{\mathrm{9}} =\:\alpha^{\mathrm{9}} −\beta^{\mathrm{9}} \\ $$$$\mathrm{we}\:\mathrm{find}\:\frac{{a}_{\mathrm{10}} −\mathrm{2}{a}_{\mathrm{8}} }{\mathrm{2}{a}_{\mathrm{9}} }\:=\:\frac{\alpha^{\mathrm{10}} −\beta^{\mathrm{10}} −\mathrm{2}\left(\alpha^{\mathrm{8}} −\beta^{\mathrm{8}} \right)}{\mathrm{2}\left(\alpha^{\mathrm{9}} −\beta^{\mathrm{9}} \right)} \\ $$$$\:=\:\frac{\alpha^{\mathrm{10}} −\beta^{\mathrm{10}} −\left(\alpha^{\mathrm{10}} −\beta^{\mathrm{10}} −\mathrm{6}\alpha^{\mathrm{9}} +\mathrm{6}\beta^{\mathrm{9}} \right)}{\mathrm{2}\left(\alpha^{\mathrm{9}} −\beta^{\mathrm{9}} \right)}=\:\mathrm{3} \\ $$

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