Let `alpha` and `beta` be the roots of `x^2-6x-2=0` with `alpha>beta` if `a_n=alpha^n-beta^n` for ` n>=1` then the value of `(a_10 - 2a_8)/(2a_9)`

Options

A. 6
B. `-6`
C. 3
D. `-3`

Answer

C

Explanation

Given, `alpha and beta` are the roots of the equation `x^(2)x6x-2=0.`
`therefore a_(n)=alpha^(n)-beta^(n)"for" n ge 1`
`alpha_(10)=alpha^(10)-beta^(10)`
`alpha_(8)=alpha^(8)-beta(8)`
`alpha+_(9)=alpha^(9)-beta^(9)`
Now, consider
`(alpha_(10)-2alpha_(8))/(2alpha_(9))=(alpha^(10)-beta^(10)-2(alpha^(8)-beta^(8)))/(2(alpha^(9)-alpha^(9)))=(alpha^(8)(alpha^(2)-2)-beta^(8)(beta^(2)-2))/(2(alpha^(9)beta^(9)))`
`=(alpha^(8).6 alpha-beta^(8)6 beta)/(2(alpha^(9)-beta^(9)))=(6alpha^(9)-6beta^(9))/(2(alpha^(9)-6beta^(9)))=6/2=3`
`[{:(,thereforealphaand beta"are the roots of"),(,x^(2)-6x-2=0orx^(2)=6x+2),(,impliesalpha^(2)=6alpha+2impliesalpha^(2)-2=6alpha),(,and beta^(2)=6beta+2impliesbeta^(2)-2=6beta):}]`
Alternate Solution
Since,`alpha and beta` are the roots of the equation
`x^(2)-6x-2=0.`
or `x^(2)=6x+2`
`therefore alpha^(2)=6alpha+2`
`impliesalpha^(10)=6alpha^(9)+2alpha^(8)" "...(i)`
Similarly, `beta^(10)=6beta^(9)+2beta^(8)" "...(ii)`
On subtracing Eq. (ii) from Eq. (i) we get
`alpha^(10)-beta^(10)=6(alpha^(9)-beta^(9))+2(alpha^(8)-beta^(8))(becausea_(n)=alpha^(n)-beta^(n))`
`impliesalpha_(10)-2alpha_(8)=6alpha_(9)implies(alpha_(10)-2alpha_(8))/(2alpha_(9))=3`