n2+3n

2

(n∈N*):

an=Sn-Sn- 1=

n2+3n

2

-

(n- 1)2+3

2

= n+ 1 (n ≥ 2)...(2 points)

∴an-an- 1=(n+ 1)-[(n- 1)+ 1]= 1(n≥2)

And a 1=S 1=2 accords with an=n+ 1.

∴{an} is arithmetic progression, the first term is 2, and the tolerance is 1.

∴ An = n+ 1 (n ∈ n *)...(4 points)

Let the common ratio of {bn} be q, then there is

b4+b5

b 1+b2

=

b 1q3+b 1q4

b 1+b 1

=q3=8

∴ Q = 2...(6 points)

And b1+B2 = b1+b1q = 3.

∴b 1= 1

∴ BN = 2n- 1...(8 points)

∴t2n=(b 1+b3+b5+…+b2n- 1)+(a2+a4+a6+…+a2n)

=( 1+22+24+…22n-2)+[3+5+7+…+(2n+ 1)]

=

1-4n

1-4

+

n(3+2n+ 1)

2

=

4n- 1

three

+N2+2n...( 12 points)