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

a(n+1)=√[bn*b(n+1)]2bn=an+an+12bn=√[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*b2b2=(a2)^2/b1=4。5√b2=√(9/2)d=√(9/2)-√2√bn=(n-1)(√(9/2)-√2)+√2得bn=(n+1)^2/2an=√bn*b(n+1)=(n+1)(n+2)/2 3。1/an=1/[(n+1)(n+2)/2]=2/(n+1)(n+2)