(1) from a1a3+2A4a4+a3a5 = (a2+a4) 2 =100, an > 0, (n∈N*) we know that a2+a4= 10, 4 is. 1.

(2) By an = 2n? 1, bn = an2+log2an = 4n-1+(n-1) is known, and the sum of the first n items of the sequence {bn} can be obtained by using the grouping summation method.

Solution: Solution: (1) Let the common ratio {an} of geometric series be q, then an=a 1qn- 1.

By the known a1a3+2a4a4+a3a5 = (a2+a4) 2 =100,

an > 0，(n∈N*)，

∴a2+a4= 10，

∵4 is the equal ratio median of a2 and a4,

∴a2a4=42= 16，

A2 and a4 are two roots of the equation x2- 10x+ 16=0.

∵q> 1,∴a2=2,a4=8，

∴

a 1q=2 a 1q3=8

The solution is a 1= 1, and q=2.

∴an=2n？ 1.

(2)∫an = 2n？ 1,

∴ bn = an2+log2an =4n- 1+(n- 1),

The sum of the first n terms of the sequence {bn}

sn =( 1+4+42+…+4n- 1)+( 1+2+3+…+n- 1)= 4n？ 1/3 +n(n？ 1) /2