题目

设各项均为正数的数列{an}和{bn}满足5^[an ],5^[bn] ,5^[a(n+1)] 成等比数列
设各项均为正数的数列{an}和{bn}满足5^[an(n为下标)],5^[bn(n为下标)] ,5^[a(n+1)(n+1为下标)] 成等比数列,lg[bn(N为下标)],lg[a(n+1)(N+1为下标)],lg[b(n+1)(n+1为下标)]成等差数列,且a1=1,b1=2,a2=3,求通项an、bn.

提问时间：2020-08-07

答案

由于:
5^[an],5^[bn],5^[a(n+1)]成等比数列
则有:
{5^[bn]}^2=5^[an]*5^[a(n+1)]
5^[bn^2]=5^[an+a(n+1)]
则:
2bn=an+a(n+1) -----(1)
由于:
lg[bn],lg[a(n+1)],lg[b(n+1)]成等差数列
则有:
2lg[a(n+1)]=lg[bn]+lg[b(n+1)]
lg[a(n+1)^2]=lg[bn*b(n+1)]
则:
[a(n+1)]^2=bn*b(n+1) -----(2)
则:
[an]=b(n-1)*bn -----(3)
由于数列{an}和{bn}各项均为正数
则由(2)(3)得:
a(n+1)=根号[bn*b(n+1)]
an=根号[b(n-1)*bn]
将以上两式代入(1)得:
2bn=根号[bn*b(n+1)]+根号[b(n-1)*bn]
2[根号bn]^2=根号[bn*b(n+1)]+根号[b(n-1)*bn]
2根号[bn]=根号[b(n+1)]+根号[b(n-1)]
设cn=根号[bn]
则有:
2cn=c(n+1)+c(n-1)
c(n+1)-cn=cn-c(n-1)
[c(n+1)-cn]/[cn-c(n-1)]=1
则:
{c(n+1)-cn}为公比为1的等比数列
则:
c(n+1)-cn=[c2-c1]=根号2/2
则:
cn
=c1+(n-1)(根号2/2)
=(n+1)/根号2
则:
bn
=(cn)^2
=(n+1)^2/2
则:
an
=根号[b(n-1)*bn]
=[n(n+1)]/2