大一高数极限问题,大哥大姐帮帮忙吧已知F﹙n﹚=1/√5﹛[﹙1-√5﹚/2]∧﹙n+1﹚-[﹙1+√5﹚/2]∧﹙n+1﹚﹜,求证当n→∝时,limF﹙n﹚/F(n+1)=﹙√5﹣1﹚/2
来源:学生作业帮助网 编辑:作业帮 时间:2024/06/21 19:26:33
![大一高数极限问题,大哥大姐帮帮忙吧已知F﹙n﹚=1/√5﹛[﹙1-√5﹚/2]∧﹙n+1﹚-[﹙1+√5﹚/2]∧﹙n+1﹚﹜,求证当n→∝时,limF﹙n﹚/F(n+1)=﹙√5﹣1﹚/2](/uploads/image/z/14330249-17-9.jpg?t=%E5%A4%A7%E4%B8%80%E9%AB%98%E6%95%B0%E6%9E%81%E9%99%90%E9%97%AE%E9%A2%98%2C%E5%A4%A7%E5%93%A5%E5%A4%A7%E5%A7%90%E5%B8%AE%E5%B8%AE%E5%BF%99%E5%90%A7%E5%B7%B2%E7%9F%A5F%EF%B9%99n%EF%B9%9A%3D1%2F%E2%88%9A5%EF%B9%9B%5B%EF%B9%991-%E2%88%9A5%EF%B9%9A%EF%BC%8F2%5D%E2%88%A7%EF%B9%99n%2B1%EF%B9%9A-%5B%EF%B9%991%2B%E2%88%9A5%EF%B9%9A%2F2%5D%E2%88%A7%EF%B9%99n%2B1%EF%B9%9A%EF%B9%9C%2C%E6%B1%82%E8%AF%81%E5%BD%93n%E2%86%92%E2%88%9D%E6%97%B6%2ClimF%EF%B9%99n%EF%B9%9A%2FF%EF%BC%88n%2B1%EF%BC%89%3D%EF%B9%99%E2%88%9A5%EF%B9%A31%EF%B9%9A%EF%BC%8F2)
大一高数极限问题,大哥大姐帮帮忙吧已知F﹙n﹚=1/√5﹛[﹙1-√5﹚/2]∧﹙n+1﹚-[﹙1+√5﹚/2]∧﹙n+1﹚﹜,求证当n→∝时,limF﹙n﹚/F(n+1)=﹙√5﹣1﹚/2
大一高数极限问题,大哥大姐帮帮忙吧
已知F﹙n﹚=1/√5﹛[﹙1-√5﹚/2]∧﹙n+1﹚-[﹙1+√5﹚/2]∧﹙n+1﹚﹜,求证当n→∝时,limF﹙n﹚/F(n+1)=﹙√5﹣1﹚/2
大一高数极限问题,大哥大姐帮帮忙吧已知F﹙n﹚=1/√5﹛[﹙1-√5﹚/2]∧﹙n+1﹚-[﹙1+√5﹚/2]∧﹙n+1﹚﹜,求证当n→∝时,limF﹙n﹚/F(n+1)=﹙√5﹣1﹚/2
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
支持 楼上
这个是斐波那契数列的通项公式,要用到线性代数的知识,那上面有,仔细看。祝你成功!!!