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)=sqr[bn*b(n+1)]
an=sqr[b(n-1)*bn]
将以上两式代入(1)得:
2bn=sqr[bn*b(n+1)]+sqr[b(n-1)*bn]
2[sqrbn]^2=sqr[bn*b(n+1)]+sqr[b(n-1)*bn]
2sqr[bn]=sqr[b(n+1)]+sqr[b(n-1)]
设cn=sqr[bn]
则有:
2cn=c(n+1)+c(n-1)
c(n+1)-cn=cn-c(n-1)=.=c1-c1
c(n+1)-cn=[c2-c1]=sqr(2)/2
c2-c1=sqr(2)/2
c3-c2=sqr(2)/2
c4-c3=sqr(2)/2
.
c(n+1)-cn=qr(2)/2
把上式累加得
则:cn=c1+(n-1)(sqr(2)/2)因为（以c1=sqr(b1)=sqr(2))
=(n+1)/sqr(2)
则:bn=(cn)^2=(n+1)^2/2
则:an=sqr[b(n-1)*bn]=[n(n+1)]/2