If #alpha,beta# are roots of #x^2 -6x-2=0# and #alpha>beta# , if #a_n =(alpha)^n - (beta)^n# , #n geq 1# then #(a_10 - 2a_8)/(2a_9)?#

Question

17150 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-03-07T16:14:04

#3#

#alpha# and #beta# are roots of #x^2-6x-2#

So, #=>alpha^2-6alpha-2=0#

#=>alpha^2=6alpha+2#

Multiply #alpha^8# both sides.

#=>alpha^2×color (blue)(alpha^8)=color (blue)(alpha^8) (6alpha+2)#

#=>alpha^10=6alpha^9+2alpha^8" equation"1#

Similarly
#=>beta^2=6beta+2#

#=>beta^10=6beta^9+2beta^8" " "equation"2#

We have
#=>a_n=alpha^n-beta^n" "("where " alpha>beta)#

we have to find
#=>(a_10-2a_8)/a_9#

Or
#=>(alpha^10-beta^10-2(alpha^8-beta^8))/(2a_9)#

From equation #1" and "2#

#=>(6alpha^9+cancel (2alpha^8)-6beta^9-cancel (2beta^8)-cancel (2alpha^8)+cancel (2beta^8))/(2a_9)#

#=>(6 (alpha^9-beta^9))/(2a_9)#

#=>(6a_9)/(2a_9)#

#=>3#