有若干个数,a1,a2,a3……an,若a1=负二分之一,从第二个数起,每个数都等于"1与它前面的那个数差的倒数",即a2=1/(1-a1),……an=1/[1-a(n-1)]求(a10·a11·a12)/(a1)
来源:学生作业帮助网 编辑:作业帮 时间:2024/06/29 01:15:45
![有若干个数,a1,a2,a3……an,若a1=负二分之一,从第二个数起,每个数都等于](/uploads/image/z/305469-45-9.jpg?t=%E6%9C%89%E8%8B%A5%E5%B9%B2%E4%B8%AA%E6%95%B0%2Ca1%2Ca2%2Ca3%E2%80%A6%E2%80%A6an%2C%E8%8B%A5a1%3D%E8%B4%9F%E4%BA%8C%E5%88%86%E4%B9%8B%E4%B8%80%2C%E4%BB%8E%E7%AC%AC%E4%BA%8C%E4%B8%AA%E6%95%B0%E8%B5%B7%2C%E6%AF%8F%E4%B8%AA%E6%95%B0%E9%83%BD%E7%AD%89%E4%BA%8E%221%E4%B8%8E%E5%AE%83%E5%89%8D%E9%9D%A2%E7%9A%84%E9%82%A3%E4%B8%AA%E6%95%B0%E5%B7%AE%E7%9A%84%E5%80%92%E6%95%B0%22%2C%E5%8D%B3a2%3D1%2F%EF%BC%881-a1%EF%BC%89%2C%E2%80%A6%E2%80%A6an%3D1%2F%5B1-a%EF%BC%88n-1%EF%BC%89%5D%E6%B1%82%EF%BC%88a10%C2%B7a11%C2%B7a12%EF%BC%89%2F%EF%BC%88a1%EF%BC%89)
有若干个数,a1,a2,a3……an,若a1=负二分之一,从第二个数起,每个数都等于"1与它前面的那个数差的倒数",即a2=1/(1-a1),……an=1/[1-a(n-1)]求(a10·a11·a12)/(a1)
有若干个数,a1,a2,a3……an,若a1=负二分之一,从第二个数起,每个数都等于"1与它前面的那个数差的倒数",即a2=1/(1-a1),……an=1/[1-a(n-1)]
求(a10·a11·a12)/(a1)
有若干个数,a1,a2,a3……an,若a1=负二分之一,从第二个数起,每个数都等于"1与它前面的那个数差的倒数",即a2=1/(1-a1),……an=1/[1-a(n-1)]求(a10·a11·a12)/(a1)
a2=1/(1-a1)=1/[1-(-1/2)]=2/3
a3=1/(1-a2)=1/(1-2/3)=3
a4=1/(1-a3)=1/(1-3)=-1/2=a1
规律:数列从第1项开始,按-1/2,2/3,3循环,每3项循环一次.
a10·a11·a12/a1
=a(3·3+1)·a(3·3+2)·a(3·3+3)/a1
=a1·a2·a3/a1
=a2·a3
=(2/3)·3
=2