Meowth 寫到:you can do it too! consider Bn= Fn / 2^(n+1)) , and Fn are the sum of 2 等比 series.
you mean this?
Fn = 1/sqrt5[(1+sqrt5/2)]^n - 1/sqrt5[(1-sqrt5/2)]^n
Let An =Summation [(1+sqrt5/2)]^n /2^(n+1)sqrt5
= 1/2sqrt5/[1- (1+sqrt5/4)]
= 2/sqrt5(3-sqrt5)
Bn = Summation [(1-sqrt5/2)]^n/2^(n+1)sqrt5
= 1/2sqrt5/[1-(1-sqrt5/4)]
= 2/sqrt5(3+sqrt5)
Answer = sum of these two.
=2/sqrt5 (6/4)
=3/sqrt5 ?