1.

if `a_(n=(alpha^(n)-beta^(n))/(alpha-beta)` where `alpha` and `beta` are roots of equation `x^(2)-x-1=0` and `b_(n)=a_(n+1)+a_(n-1)` thenA. `b_(n)=alpha^(n)+beta^(n)`B. `underset(n=1)overset(infty)sum(b_(n))/(10^(n))=(8)/(89)`C. `underset(n=1)overset(infty)sum(a_(n))/(10^(n))=(10)/(89)`D. `a_(1)+a_(2)+…a_(n)=a_(n+2)-1`

Answer» Correct Answer - A::C::D
(A). `b_(n)=a_(n+1)+a_(n-1)-(alpha^(n+1)-beta^(n+_1))/(alpha-beta)+(alpha^(n-1)-beta^(n-1))/(alpha-beta)=(alpha^(n-1)(alpha^(2)+1)-beta^(n-1)(beta^(2)+1))/(alpha-beta)`
`=(alpha^(n-1)(alpha+2)-beta^(n-1)(beta+2))/(alpha-beta)=(alpha^(n-1)((5+sqrt(5))/(2))-beta^(n-1)((5-sqrt(5))/(2)))/(alpha-beta)`
`=(sqrt(5)alpha^n-1)((sqrt(5)+1)/(2))-sqrt(5)beta^(n-1)((sqrt(5)-1)/(2)))/(alpha-beta)=(sqrt(5)(alpha^(n)+beta^(n)))/(alpha-beta)=alpha^(n)+beta^(n)" "becausealpha-beta`
(B). `underset(n=1)overset(infty)sum(b_(0))/(10^(n))=sum((alpha)/(10))^(n)+sum((beta)/(10))^(n)+sum((beta)/(10))^(n)=((alpha)/(10))/(1-(alpha)/(10))+((beta)/(10))/(1-(beta)/(10))=(alpha)/(10-alpha)+(beta)/(10-beta)`
`=(10(alpha+beta)-2alphabeta)/(100-10(alpha+beta)+alphabeta)=(10+2)/(89)=(12)/(89)`
(C). `underset(n=1)overset(infty)(a_(n))/(10^(n))=sum(alpha^(n)-beta^(n))/((alpha-beta)10^(n))=(1)/(alpha-beta)(((alpha)/(10))/(1-(alpha)/(10))-((beta)/(10))/(1-(beta)/(10)))(1)/(alpha-beta)((alpha)/(10-alpha)-(beta)/(10-beta))`
`=(1)/(alpha-beta.((10(alpha-beta)-alphabeta+alphabeta))/(100-10(alpha+beta)+alphabeta))=(10)/(89)`
(D). `a_(1)+a_(2)+...a_(n)=suma_(i)=(sumalpha^(n)-sumbeta^(i))/(alpha-beta)=((alpha(1-alpha^(n)))/((1-alpha))-(beta(1-beta^(n)))/((1-beta)))/(alpha-beta)`
`=((alpha+1)(1+alpha^(n))-(beta+1)(1-beta^(n)))/((1-alpha)(1-beta)(alpha-beta))=(alpha^(2)-alpha^(n+2)-beta^(2)+beta^(n+2))/((1-alpha)(1-beta)(alpha-beta))=(sqrt(5)+beta^(n+2)-alpha^(n+2))/(beta-alpha)=-1+a_(n+2)`


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