(1)
由B1+B3=5,B1B3=4可得方程组
a1+a1q^2 = 5 => a1(1+q^2) = 5 ①
a1^2q^2 = 4 ②
∵Bn是递增数列
∴{a1>0且q>1} 或者 {a11 => a1q = 2 ③
联立方程①、③得,a1 = 1,q =2
∴Bn = 2^(n-1) (n属于正实数)
(2)证明：
An = log2(2^(n-1)) + 3 = n-1 + 3 = n+2
对于任意n∈N+
A(n+1) - An = (n+1+2) - (n+2) = 1为常数
∴数列{An}是等差数列
(3) A1^2+A2+A3+.Am