Question

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

Answer 1

1.

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)