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)