正数列{an}和{bn}满足对任意自然数n,an,bn,an+1成等差数列,bn,an+1,bn+1成等比数列

来源：学生作业帮编辑：作业帮分类：数学作业时间：2024/04/29 18:35:29

正数列{an}和{bn}满足对任意自然数n,an,bn,an+1成等差数列,bn,an+1,bn+1成等比数列
（1）证明：数列{√bn}成等差数列
（2）若a1=1,b1=2,a2=3,求数列{an},{bn}的通项公式
（3）在（2）的前提下求{1/an}的通项公式

a(n+1)=√[bn*b(n+1)]
2bn=an+an+1
2bn=√[bn*b(n-1)]+√[bn*b(n+1)]
2√bn=√b(n-1)+√b(n+1)
所以数列{√bn}为等差数列
2.
√b1=√2
(a2)^2=b1*b2
b2=(a2)^2/b1=4.5
√b2=√(9/2)
d=√(9/2)-√2
√bn=(n-1)(√(9/2)-√2)+√2
得bn=(n+1)^2/2
an=√bn*b(n+1)=(n+1)(n+2)/2
3.
1/an=1/[(n+1)(n+2)/2]=2/(n+1)(n+2)

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