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)