Question

Let `alpha and beta` be the roots of `x^2-x-1=0`, with `alpha gt beta`. For all positive integers n, define
`a_n=(alpha^n=beta^n)/(alpha-beta),nge 1`.
`b_1=1 and b_b=a_(n-1)+a_n+1,n ge 2`
Then which of the following options is/are correct ?
A. `sum_(n=1)^(oo)(b_n)/(10^n)=(8)/(89)`
B. `b_n=alpha^n+beta^n " for all " n ge 1`
C. `a_1+a_2+a_3+.......+a_n=a_(n+2)-1" for all " n ge 1`
D. `sum_(n=1)^(oo)(a_n)/(10^n)=(10)/(89)`

Answer 1

Correct Answer - B::C::D
Given quadratic equation `x^2-x-1=0` having roots `alpha` and `beta,(alpha gt beta)`
So, `alpha=(1+sqrt(5))/(2)and beta =(1-sqrt(5))/(2)`
and `alpha +beta =1,alpha beta =-1`
`because a_n=(a^n-beta^n)/(alpha-beta),n ge1`
So,
`a_(n+1)=((alpha^n+1)-beta^(n+1))/(alpha-beta)=alpha^n+alpha^n-1beta+alpha^n-2beta+.......alphabeta^(n-1)+beta^(n)`
` =alpha^n-alpha^n-1-alpha^n-3beta......-beta ^n-2+beta^n`
`=alpha^n+beta^n-(alpha^n-1+alpha^n-3beta+....+beta^n-2)`
`=alpha^n+beta^n-a_(n-1)`
`[as a_(n-1)=(alph^n-1 -beta^n-1)/(alpha-beta)=alpha^n-2+alpha^n-3beta+....beta^n-2]`
`impliesalpha_(n+1)+alpha_(n-1)=alpha^(n)+beta^(n)=b_(n),AAnge1`
So,option (b) is incorrect
Now `Sigma_(n=1)^(oo) (b_n)/(10^n)=Sigma_(n=1)^(oo) (a^n+b^n)/(10^n)`
` =Sigma_(n=1)^(oo) ((alpha)/(10))^n+Sigma_(n=1)^(oo) ((beta)/(10))^n [because |(alpha)/(10)|lt 1 and |(beta)/(10)| lt 1 and |(beta)/(10)lt 1|`
` =((alpha)/(10))/(10(alpha)/(10))+((beta)/(10))/(1-(beta)/(10))=(alpha)/(10-alpha)+(beta)/(1-beta)`
` =(10alpha-alpha beta +10beta-alpha beta)/((10-alpha)(10-beta))=(10(alpha+beta)-2alphabeta)/(100-10(alpha+beta)+alpha beta)`
`=(10(1)-2(-1))/(100-100(1)-1)`
`=(12)/(89)`
So, option (a) is not correct.
`because alpha^2=alpha +1 and beta^2=beta+1`
`rArr alpha^n+1=alpha^n+1+alpha^n and beta^n+2=beta^n+1+beta^n`.
`rArr (alpha^n+2+beta^n+2)=(alpha^n+1+beta^n+1)+(alpha^n+beta^n)`
`rArr a_(n+1)=a_(n+1)+a_n`
Similarly `a_(n+1)=a_n+a_n-1`
`a_n=a_n-1+a_(n-2)`
`.............`
`...................`
On adding , we get
`a_(n+2)=(a_n+a_n-1+a_n-2+....+a_2+a_1)+a_2`
`[because a_2=(alpha^2-beta^2)/(alpha-beta)=alpha+beta=1]`
So, ` a_n+2-1=a_1=a_1+a_2+a_3+......+a_n`
So, option (c) is aslo correct.
And, now `Sigma_(n =1)^(oo) (a_n)/(10^n)=Sigma_(n=1)^(oo) (alpha^n-beta^n)/((alpha-beta) 10^n)`
` =(1)/(alpha-beta)[Sigma _(n=1)^(oo) ((alpha)/(10))^n-Sigma _(n=1)^(oo) ((beta)/(10))^n]`
` =(1)/(alpha-beta) [((alpha)/(10))/(1-(alpha)/(10))-((beta)/(10))/(1-(beta)/(10))],[as |(alpha)/(10)|lt 1 and |(beta)/(10)|lt 1]`
`=(1)/(alpha-beta) ((alpha)/(10-alpha)-(beta)/(10-beta))=(1)/(alpha-beta)[(10alpha-alphabeta-10beta+alpha beta)/(100-10(alpha +beta)+alpha beta)]`
` =(10(alpha -beta))/((alpha -beta)[100-10(alpha+beta )+alpha beta])=(10)/(100-10-1)=(10)/(89)`
Hence options, (b),(c) and (d) are correct.