Question and Answers Forum

All Questions      Topic List

Relation and Functions Questions

Previous in All Question      Next in All Question      

Previous in Relation and Functions      Next in Relation and Functions      

Question Number 148502 by mathmax by abdo last updated on 28/Jul/21

let α and β roots of x^2 +x+2 simplify Σ_(k=0) ^(n−1) (α^k +β^k ) and Σ_(k=0) ^(n−1) ( (1/α^k )+(1/β^k ))

$$\mathrm{let}\:\alpha\:\mathrm{and}\:\beta\:\mathrm{roots}\:\mathrm{of}\:\mathrm{x}^{\mathrm{2}} +\mathrm{x}+\mathrm{2} \\ $$$$\mathrm{simplify}\:\:\sum_{\mathrm{k}=\mathrm{0}} ^{\mathrm{n}−\mathrm{1}} \:\:\left(\alpha^{\mathrm{k}} \:+\beta^{\mathrm{k}} \right)\:\:\mathrm{and}\:\sum_{\mathrm{k}=\mathrm{0}} ^{\mathrm{n}−\mathrm{1}} \left(\:\frac{\mathrm{1}}{\alpha^{\mathrm{k}} }+\frac{\mathrm{1}}{\beta^{\mathrm{k}} }\right) \\ $$

Answered by ArielVyny last updated on 28/Jul/21

x^2 +x+2=0→(x+(1/2))^2 −(1/4)+(8/4)=0 →(x+(1/2))^2 −(i((√7)/2))^2 =0→(x+(1/2)+i((√7)/2))(x+(1/2)−i((√7)/2))=0 α=−((1+i(√7))/2) and β=((−1+i(√7))/2) ∣α∣=∣β∣(√((−(1/2))^2 +(((√7)/2))^2 ))=(√((1/4)+(7/4)))=((2(√2))/2)=(√2) α=(√2)(−(1/(2(√2)))−i((√7)/(2(√2))))=(√2)(−((√2)/4)−i((√(14))/4))=(√2)e^(iθ) β=(√2)(−((√2)/4)+((√(14))/4))=(√2)e^(−iθ) θ=110,7048111∼ U_n =Σ_(k=0) ^(n−1) (α^k +β^k )=Σ_(k=0) ^(n−1) α^k +Σ_(k=0) ^(n−1) β^k Σ_(k=0) ^(n−1) (e^(iθ) )^k +Σ_(k=0) ^(n−1) (e^(−iθ) )^k ((1−(e^(iθ) )^n )/(1−e^(iθ) ))+((1−(e^(−iθ) )^n )/(1−e^(−iθ) ))=((1−e^(inθ) )/(1−e^(iθ) ))+((1−(1/e^(inθ) ))/(1−(1/e^(iθ) ))) =((1−e^(inθ) )/(1−e^(iθ) ))+(((e^(inθ) −1)/e^(inθ) )/((e^(iθ) −1)/e^(iθ) ))=((1−e^(inθ) )/(1−e^(iθ) ))−((e^(iθ(n+1)) −e^(iθ) )/e^(inθ) )(1/(1−e^(iθ) )) U_n =(1/(1−e^(iθ) ))(1−e^(inθ) −((e^(iθ(n+1)) −e^(iθ) )/e^(inθ) )) U_n =(1/(1−e^(iθ) ))(1−e^(inθ) −e^(iθ) −e^(iθ(1−n)) ) U_n =(1/(1−e^(iθ) ))(1−e^(inθ) −e^(iθ) (1+e^(−in) ))

$${x}^{\mathrm{2}} +{x}+\mathrm{2}=\mathrm{0}\rightarrow\left({x}+\frac{\mathrm{1}}{\mathrm{2}}\right)^{\mathrm{2}} −\frac{\mathrm{1}}{\mathrm{4}}+\frac{\mathrm{8}}{\mathrm{4}}=\mathrm{0} \\ $$$$\rightarrow\left({x}+\frac{\mathrm{1}}{\mathrm{2}}\right)^{\mathrm{2}} −\left({i}\frac{\sqrt{\mathrm{7}}}{\mathrm{2}}\right)^{\mathrm{2}} =\mathrm{0}\rightarrow\left({x}+\frac{\mathrm{1}}{\mathrm{2}}+{i}\frac{\sqrt{\mathrm{7}}}{\mathrm{2}}\right)\left({x}+\frac{\mathrm{1}}{\mathrm{2}}−{i}\frac{\sqrt{\mathrm{7}}}{\mathrm{2}}\right)=\mathrm{0} \\ $$$$\alpha=−\frac{\mathrm{1}+{i}\sqrt{\mathrm{7}}}{\mathrm{2}}\:{and}\:\beta=\frac{−\mathrm{1}+{i}\sqrt{\mathrm{7}}}{\mathrm{2}} \\ $$$$\mid\alpha\mid=\mid\beta\mid\sqrt{\left(−\frac{\mathrm{1}}{\mathrm{2}}\right)^{\mathrm{2}} +\left(\frac{\sqrt{\mathrm{7}}}{\mathrm{2}}\right)^{\mathrm{2}} }=\sqrt{\frac{\mathrm{1}}{\mathrm{4}}+\frac{\mathrm{7}}{\mathrm{4}}}=\frac{\mathrm{2}\sqrt{\mathrm{2}}}{\mathrm{2}}=\sqrt{\mathrm{2}} \\ $$$$\alpha=\sqrt{\mathrm{2}}\left(−\frac{\mathrm{1}}{\mathrm{2}\sqrt{\mathrm{2}}}−{i}\frac{\sqrt{\mathrm{7}}}{\mathrm{2}\sqrt{\mathrm{2}}}\right)=\sqrt{\mathrm{2}}\left(−\frac{\sqrt{\mathrm{2}}}{\mathrm{4}}−{i}\frac{\sqrt{\mathrm{14}}}{\mathrm{4}}\right)=\sqrt{\mathrm{2}}{e}^{{i}\theta} \\ $$$$\beta=\sqrt{\mathrm{2}}\left(−\frac{\sqrt{\mathrm{2}}}{\mathrm{4}}+\frac{\sqrt{\mathrm{14}}}{\mathrm{4}}\right)=\sqrt{\mathrm{2}}{e}^{−{i}\theta} \\ $$$$\theta=\mathrm{110},\mathrm{7048111}\sim \\ $$$${U}_{{n}} =\sum_{{k}=\mathrm{0}} ^{{n}−\mathrm{1}} \left(\alpha^{{k}} +\beta^{{k}} \right)=\underset{{k}=\mathrm{0}} {\overset{{n}−\mathrm{1}} {\sum}}\alpha^{{k}} +\underset{{k}=\mathrm{0}} {\overset{{n}−\mathrm{1}} {\sum}}\beta^{{k}} \\ $$$$\underset{{k}=\mathrm{0}} {\overset{{n}−\mathrm{1}} {\sum}}\left({e}^{{i}\theta} \right)^{{k}} +\underset{{k}=\mathrm{0}} {\overset{{n}−\mathrm{1}} {\sum}}\left({e}^{−{i}\theta} \right)^{{k}} \\ $$$$\frac{\mathrm{1}−\left({e}^{{i}\theta} \right)^{{n}} }{\mathrm{1}−{e}^{{i}\theta} }+\frac{\mathrm{1}−\left({e}^{−{i}\theta} \right)^{{n}} }{\mathrm{1}−{e}^{−{i}\theta} }=\frac{\mathrm{1}−{e}^{{in}\theta} }{\mathrm{1}−{e}^{{i}\theta} }+\frac{\mathrm{1}−\frac{\mathrm{1}}{{e}^{{in}\theta} }}{\mathrm{1}−\frac{\mathrm{1}}{{e}^{{i}\theta} }} \\ $$$$=\frac{\mathrm{1}−{e}^{{in}\theta} }{\mathrm{1}−{e}^{{i}\theta} }+\frac{\frac{{e}^{{in}\theta} −\mathrm{1}}{{e}^{{in}\theta} }}{\frac{{e}^{{i}\theta} −\mathrm{1}}{{e}^{{i}\theta} }}=\frac{\mathrm{1}−{e}^{{in}\theta} }{\mathrm{1}−{e}^{{i}\theta} }−\frac{{e}^{{i}\theta\left({n}+\mathrm{1}\right)} −{e}^{{i}\theta} }{{e}^{{in}\theta} }\frac{\mathrm{1}}{\mathrm{1}−{e}^{{i}\theta} } \\ $$$${U}_{{n}} =\frac{\mathrm{1}}{\mathrm{1}−{e}^{{i}\theta} }\left(\mathrm{1}−{e}^{{in}\theta} −\frac{{e}^{{i}\theta\left({n}+\mathrm{1}\right)} −{e}^{{i}\theta} }{{e}^{{in}\theta} }\right) \\ $$$${U}_{{n}} =\frac{\mathrm{1}}{\mathrm{1}−{e}^{{i}\theta} }\left(\mathrm{1}−{e}^{{in}\theta} −{e}^{{i}\theta} −{e}^{{i}\theta\left(\mathrm{1}−{n}\right)} \right) \\ $$$${U}_{{n}} =\frac{\mathrm{1}}{\mathrm{1}−{e}^{{i}\theta} }\left(\mathrm{1}−{e}^{{in}\theta} −{e}^{{i}\theta} \left(\mathrm{1}+{e}^{−{in}} \right)\right) \\ $$

Answered by Olaf_Thorendsen last updated on 28/Jul/21

Let u_0 = 2, u_1 = −1 and ∀n ≥ 2, u_n = −u_(n−1) −2u_(n−2) We solve r^2 +r+2 = 0. The roots are noted α and β. Then u_n = λα^n +μβ^n . With u_0 and u_1 , λ = μ = 1. ⇒ ∀n∈N, u_n = α^n +β^n S_(n+2) = Σ_(k=0) ^(n+2) (α^k +β^k ) = Σ_(k=0) ^(n+2) u_k S_(n+2) = u_0 +u_1 +Σ_(k=2) ^(n+2) u_k S_(n+2) = u_0 +u_1 +Σ_(k=0) ^n u_(k+2) S_(n+2) = u_0 +u_1 −Σ_(k=0) ^n u_(k+1) −2Σ_(k=0) ^n u_k S_(n+2) = u_0 +u_1 −(S_(n+2) −u_0 −u_(n+2) ) −2(S_(n+2) −u_(n+1) −u_(n+2) ) 4S_(n+2) = 3u_(n+2) +2u_(n+1) +2u_0 +u_1 S_(n+2) = (3/4)(α^(n+2) +β^(n+2) )+(1/2)(α^(n+1) +β^(n+1) )+(3/4) Σ_(k=0) ^(n−1) (α^k +α^k ) = S_(n−1) = (3/4)(α^(n−1) +β^(n−1) )+(1/2)(α^(n−2) +β^(n−2) )+(3/4) = (1/4)α^(n−2) (3α+2)+(1/4)β^(n−2) (3β+2)+(3/4) = (1/4)α^(n−2) (2α−α^2 )+(1/4)β^(n−2) (2β−β^2 )+(3/4) Σ_(k=0) ^(n−1) (α^k +α^k ) = (1/4)α^(n−1) (2−α)+(1/4)β^(n−1) (2−β)+(3/4) α+β = −(b/a) = −1 and αβ = (c/a) = 2 (x−(1/α))(x−(1/β)) = x^2 −((1/α)+(1/β))x+(1/(αβ)) = x^2 −((α+β)/(αβ))x+(1/(αβ)) = x^2 +(1/2)x+(1/2) ⇒ ForΣ_(k=0) ^(n−1) ((1/α^k )+(1/β^k )) : same kind of calculous with u_n = −(1/2)u_(n−1) −(1/2)u_(n−2)

$$\mathrm{Let}\:{u}_{\mathrm{0}} \:=\:\mathrm{2},\:{u}_{\mathrm{1}} \:=\:−\mathrm{1} \\ $$$$\mathrm{and}\:\forall{n}\:\geqslant\:\mathrm{2},\:{u}_{{n}} \:=\:−{u}_{{n}−\mathrm{1}} −\mathrm{2}{u}_{{n}−\mathrm{2}} \\ $$$$\mathrm{We}\:\mathrm{solve}\:{r}^{\mathrm{2}} +{r}+\mathrm{2}\:=\:\mathrm{0}. \\ $$$$\mathrm{The}\:\mathrm{roots}\:\mathrm{are}\:\mathrm{noted}\:\alpha\:\mathrm{and}\:\beta. \\ $$$$\mathrm{Then}\:{u}_{{n}} =\:\lambda\alpha^{{n}} +\mu\beta^{{n}} . \\ $$$$\mathrm{With}\:{u}_{\mathrm{0}} \:\mathrm{and}\:{u}_{\mathrm{1}} ,\:\lambda\:=\:\mu\:=\:\mathrm{1}. \\ $$$$\Rightarrow\:\forall{n}\in\mathbb{N},\:{u}_{{n}} \:=\:\alpha^{{n}} +\beta^{{n}} \\ $$$$ \\ $$$$\mathrm{S}_{{n}+\mathrm{2}} \:=\:\underset{{k}=\mathrm{0}} {\overset{{n}+\mathrm{2}} {\sum}}\left(\alpha^{{k}} +\beta^{{k}} \right)\:=\:\underset{{k}=\mathrm{0}} {\overset{{n}+\mathrm{2}} {\sum}}{u}_{{k}} \\ $$$$\mathrm{S}_{{n}+\mathrm{2}} \:=\:{u}_{\mathrm{0}} +{u}_{\mathrm{1}} +\underset{{k}=\mathrm{2}} {\overset{{n}+\mathrm{2}} {\sum}}{u}_{{k}} \\ $$$$\mathrm{S}_{{n}+\mathrm{2}} \:=\:{u}_{\mathrm{0}} +{u}_{\mathrm{1}} +\underset{{k}=\mathrm{0}} {\overset{{n}} {\sum}}{u}_{{k}+\mathrm{2}} \\ $$$$\mathrm{S}_{{n}+\mathrm{2}} \:=\:{u}_{\mathrm{0}} +{u}_{\mathrm{1}} −\underset{{k}=\mathrm{0}} {\overset{{n}} {\sum}}{u}_{{k}+\mathrm{1}} −\mathrm{2}\underset{{k}=\mathrm{0}} {\overset{{n}} {\sum}}{u}_{{k}} \\ $$$$\mathrm{S}_{{n}+\mathrm{2}} \:=\:{u}_{\mathrm{0}} +{u}_{\mathrm{1}} −\left(\mathrm{S}_{{n}+\mathrm{2}} −{u}_{\mathrm{0}} −{u}_{{n}+\mathrm{2}} \right) \\ $$$$−\mathrm{2}\left(\mathrm{S}_{{n}+\mathrm{2}} −{u}_{{n}+\mathrm{1}} −{u}_{{n}+\mathrm{2}} \right) \\ $$$$ \\ $$$$\mathrm{4S}_{{n}+\mathrm{2}} \:=\:\mathrm{3}{u}_{{n}+\mathrm{2}} +\mathrm{2}{u}_{{n}+\mathrm{1}} +\mathrm{2}{u}_{\mathrm{0}} +{u}_{\mathrm{1}} \\ $$$$\mathrm{S}_{{n}+\mathrm{2}} \:=\:\frac{\mathrm{3}}{\mathrm{4}}\left(\alpha^{{n}+\mathrm{2}} +\beta^{{n}+\mathrm{2}} \right)+\frac{\mathrm{1}}{\mathrm{2}}\left(\alpha^{{n}+\mathrm{1}} +\beta^{{n}+\mathrm{1}} \right)+\frac{\mathrm{3}}{\mathrm{4}} \\ $$$$ \\ $$$$\underset{{k}=\mathrm{0}} {\overset{{n}−\mathrm{1}} {\sum}}\left(\alpha^{{k}} +\alpha^{{k}} \right)\:=\:\mathrm{S}_{{n}−\mathrm{1}} \\ $$$$=\:\frac{\mathrm{3}}{\mathrm{4}}\left(\alpha^{{n}−\mathrm{1}} +\beta^{{n}−\mathrm{1}} \right)+\frac{\mathrm{1}}{\mathrm{2}}\left(\alpha^{{n}−\mathrm{2}} +\beta^{{n}−\mathrm{2}} \right)+\frac{\mathrm{3}}{\mathrm{4}} \\ $$$$=\:\frac{\mathrm{1}}{\mathrm{4}}\alpha^{{n}−\mathrm{2}} \left(\mathrm{3}\alpha+\mathrm{2}\right)+\frac{\mathrm{1}}{\mathrm{4}}\beta^{{n}−\mathrm{2}} \left(\mathrm{3}\beta+\mathrm{2}\right)+\frac{\mathrm{3}}{\mathrm{4}} \\ $$$$=\:\frac{\mathrm{1}}{\mathrm{4}}\alpha^{{n}−\mathrm{2}} \left(\mathrm{2}\alpha−\alpha^{\mathrm{2}} \right)+\frac{\mathrm{1}}{\mathrm{4}}\beta^{{n}−\mathrm{2}} \left(\mathrm{2}\beta−\beta^{\mathrm{2}} \right)+\frac{\mathrm{3}}{\mathrm{4}} \\ $$$$ \\ $$$$\underset{{k}=\mathrm{0}} {\overset{{n}−\mathrm{1}} {\sum}}\left(\alpha^{{k}} +\alpha^{{k}} \right)\:=\:\frac{\mathrm{1}}{\mathrm{4}}\alpha^{{n}−\mathrm{1}} \left(\mathrm{2}−\alpha\right)+\frac{\mathrm{1}}{\mathrm{4}}\beta^{{n}−\mathrm{1}} \left(\mathrm{2}−\beta\right)+\frac{\mathrm{3}}{\mathrm{4}} \\ $$$$ \\ $$$$ \\ $$$$\alpha+\beta\:=\:−\frac{{b}}{{a}}\:=\:−\mathrm{1}\:\mathrm{and}\:\alpha\beta\:=\:\frac{{c}}{{a}}\:=\:\mathrm{2} \\ $$$$\left({x}−\frac{\mathrm{1}}{\alpha}\right)\left({x}−\frac{\mathrm{1}}{\beta}\right)\:=\:{x}^{\mathrm{2}} −\left(\frac{\mathrm{1}}{\alpha}+\frac{\mathrm{1}}{\beta}\right){x}+\frac{\mathrm{1}}{\alpha\beta} \\ $$$$=\:{x}^{\mathrm{2}} −\frac{\alpha+\beta}{\alpha\beta}{x}+\frac{\mathrm{1}}{\alpha\beta} \\ $$$$=\:{x}^{\mathrm{2}} +\frac{\mathrm{1}}{\mathrm{2}}{x}+\frac{\mathrm{1}}{\mathrm{2}} \\ $$$$\Rightarrow\:\mathrm{For}\underset{{k}=\mathrm{0}} {\overset{{n}−\mathrm{1}} {\sum}}\left(\frac{\mathrm{1}}{\alpha^{{k}} }+\frac{\mathrm{1}}{\beta^{{k}} }\right)\::\:\mathrm{same}\:\mathrm{kind}\:\mathrm{of}\:\mathrm{calculous} \\ $$$$\mathrm{with}\:{u}_{{n}} \:=\:−\frac{\mathrm{1}}{\mathrm{2}}{u}_{{n}−\mathrm{1}} −\frac{\mathrm{1}}{\mathrm{2}}{u}_{{n}−\mathrm{2}} \\ $$

Answered by mathmax by abdo last updated on 28/Jul/21

x^(2 ) +x+2=0 →Δ=1−8=−7 ⇒x_1 =((−1+i(√7))/2) and x_2 =((−1−i(√7))/2) ∣x_1 ∣=(1/2)(√(1+7))=((2(√2))/2) ⇒x_1 =(√2)(((−1)/(2(√2)))+((i(√7))/(2(√2))))=(√2)e^(−iarctan((√7))) ⇒x_2 =(√2)e^(iarctan((√7))) ⇒ A_n =Σ_(k=0) ^(n−1) α^k +β^k =Σ_(k=0) ^(n−1) ((√2)e^(iarctan((√7))) )^k +((√2)e^(−iarctan((√7))) )^k =Σ_(k=0) ^(n−1) ((√2))^k { e^(ikarctan((√7))) +e^(−ikarctan((√7))) } =2 Σ_(k=0) ^(n−1) ((√2))^k cos(karctan((√7))) let T_k (x)=cos(karctanx) ⇒A_n =2Σ_(k=0) ^(n−1) ((√2))^k T_k ((√7)) Σ_(k=0) ^(n−1) ((1/α^k )+(1/β^k ))=Σ_(k=0) ^(n−1) (((α^k +β^k )/((αβ)^k ))) =Σ_(k=0) ^(n−1) (1/2^k )(α^k +β^k ) =2Σ_(k=0) ^(n−1) (1/2^k )((√2))^k cos(karctan((√7))) =2Σ_(k=0) ^(n−1) ((√2))^(−k) T_k ((√7))

$$\mathrm{x}^{\mathrm{2}\:} +\mathrm{x}+\mathrm{2}=\mathrm{0}\:\rightarrow\Delta=\mathrm{1}−\mathrm{8}=−\mathrm{7}\:\Rightarrow\mathrm{x}_{\mathrm{1}} =\frac{−\mathrm{1}+\mathrm{i}\sqrt{\mathrm{7}}}{\mathrm{2}}\:\mathrm{and}\:\mathrm{x}_{\mathrm{2}} =\frac{−\mathrm{1}−\mathrm{i}\sqrt{\mathrm{7}}}{\mathrm{2}} \\ $$$$\mid\mathrm{x}_{\mathrm{1}} \mid=\frac{\mathrm{1}}{\mathrm{2}}\sqrt{\mathrm{1}+\mathrm{7}}=\frac{\mathrm{2}\sqrt{\mathrm{2}}}{\mathrm{2}}\:\Rightarrow\mathrm{x}_{\mathrm{1}} =\sqrt{\mathrm{2}}\left(\frac{−\mathrm{1}}{\mathrm{2}\sqrt{\mathrm{2}}}+\frac{\mathrm{i}\sqrt{\mathrm{7}}}{\mathrm{2}\sqrt{\mathrm{2}}}\right)=\sqrt{\mathrm{2}}\mathrm{e}^{−\mathrm{iarctan}\left(\sqrt{\mathrm{7}}\right)} \\ $$$$\Rightarrow\mathrm{x}_{\mathrm{2}} =\sqrt{\mathrm{2}}\mathrm{e}^{\mathrm{iarctan}\left(\sqrt{\mathrm{7}}\right)} \:\Rightarrow \\ $$$$\mathrm{A}_{\mathrm{n}} =\sum_{\mathrm{k}=\mathrm{0}} ^{\mathrm{n}−\mathrm{1}} \:\alpha^{\mathrm{k}} \:+\beta^{\mathrm{k}} =\sum_{\mathrm{k}=\mathrm{0}} ^{\mathrm{n}−\mathrm{1}} \:\left(\sqrt{\mathrm{2}}\mathrm{e}^{\mathrm{iarctan}\left(\sqrt{\mathrm{7}}\right)} \right)^{\mathrm{k}} \:+\left(\sqrt{\mathrm{2}}\mathrm{e}^{−\mathrm{iarctan}\left(\sqrt{\mathrm{7}}\right)} \right)^{\mathrm{k}} \\ $$$$=\sum_{\mathrm{k}=\mathrm{0}} ^{\mathrm{n}−\mathrm{1}} \:\left(\sqrt{\mathrm{2}}\right)^{\mathrm{k}} \left\{\:\mathrm{e}^{\mathrm{ikarctan}\left(\sqrt{\mathrm{7}}\right)} \:\:+\mathrm{e}^{−\mathrm{ikarctan}\left(\sqrt{\mathrm{7}}\right)} \right\} \\ $$$$=\mathrm{2}\:\sum_{\mathrm{k}=\mathrm{0}} ^{\mathrm{n}−\mathrm{1}} \:\left(\sqrt{\mathrm{2}}\right)^{\mathrm{k}} \:\mathrm{cos}\left(\mathrm{karctan}\left(\sqrt{\mathrm{7}}\right)\right) \\ $$$$\mathrm{let}\:\mathrm{T}_{\mathrm{k}} \left(\mathrm{x}\right)=\mathrm{cos}\left(\mathrm{karctanx}\right)\:\Rightarrow\mathrm{A}_{\mathrm{n}} =\mathrm{2}\sum_{\mathrm{k}=\mathrm{0}} ^{\mathrm{n}−\mathrm{1}} \left(\sqrt{\mathrm{2}}\right)^{\mathrm{k}} \:\mathrm{T}_{\mathrm{k}} \left(\sqrt{\mathrm{7}}\right) \\ $$$$\sum_{\mathrm{k}=\mathrm{0}} ^{\mathrm{n}−\mathrm{1}} \left(\frac{\mathrm{1}}{\alpha^{\mathrm{k}} }+\frac{\mathrm{1}}{\beta^{\mathrm{k}} }\right)=\sum_{\mathrm{k}=\mathrm{0}} ^{\mathrm{n}−\mathrm{1}} \left(\frac{\alpha^{\mathrm{k}} \:+\beta^{\mathrm{k}} }{\left(\alpha\beta\right)^{\mathrm{k}} }\right) \\ $$$$=\sum_{\mathrm{k}=\mathrm{0}} ^{\mathrm{n}−\mathrm{1}} \:\frac{\mathrm{1}}{\mathrm{2}^{\mathrm{k}} }\left(\alpha^{\mathrm{k}} \:+\beta^{\mathrm{k}} \right)\:=\mathrm{2}\sum_{\mathrm{k}=\mathrm{0}} ^{\mathrm{n}−\mathrm{1}} \:\frac{\mathrm{1}}{\mathrm{2}^{\mathrm{k}} }\left(\sqrt{\mathrm{2}}\right)^{\mathrm{k}} \:\mathrm{cos}\left(\mathrm{karctan}\left(\sqrt{\mathrm{7}}\right)\right) \\ $$$$=\mathrm{2}\sum_{\mathrm{k}=\mathrm{0}} ^{\mathrm{n}−\mathrm{1}} \:\left(\sqrt{\mathrm{2}}\right)^{−\mathrm{k}} \:\mathrm{T}_{\mathrm{k}} \left(\sqrt{\mathrm{7}}\right) \\ $$$$ \\ $$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com