第1个回答 2012-09-25
你好!
设 a= (1-√5)/2
1 - a = (1+√5)/2
| a/(1-a) | < 1
lim(n→+∞) [a/(1-a)]^n = 0
Fn = 1/√5 * [ a^(n+1) - (1-a)^(n+1) ]
F(n+1) = 1/√5 * [ a^(n+2) - (1-a)^(n+2) ]
n→+∞ lim Fn / F(n-1)
= n→+∞ lim [ a^(n+1) - (1-a)^(n+1) ] / [ a^(n+2) - (1-a)^(n+2) ]
= n→+∞ lim { [ a/(1-a) ]^(n+1) - 1 } / { a* [ a/(1-a) ]^(n+1) - (1-a) }
= -1 / (a-1)
= 2 / (1+√5)
= (√5 - 1) / 2本回答被提问者和网友采纳