解题思路:(1)设{an}的公差为d,由b2+S2=12,b1=1,q=
S
2
b
2
,可求得q=3,d=3,从而可得an与bn;
(2)由(1)知,an=3n,于是可得Sn=
3n(n+1)
2
,[1/Sn]=[2/3]([1/n]-[1/n+1]),通过列项相消法即可求得答案.
(1)设{an}的公差为d,
∵b2+S2=12,b1=1,q=
S2
b2,
∴
q+6+d=12
q2=6+d,解得q=3或q=-4(舍),d=3.
故an=3n,bn=3n-1…(4分)
(2)Sn=
n(3+3n)/2]=
3n(n+1)
2,∴[1
Sn=
2
3n(n+1)=
2/3]([1/n]-[1/n+1]),
∴[1
S1+
1
S2+…+
1
Sn=
2/3](1-[1/2]+[1/2]-[1/3]+…+[1/n]-[1/n+1])=[2/3](1-[1/n+1])…(8分)
∵n≥1,∴0<
点评:
本题考点: 数列的求和;等差数列的前n项和.
考点点评: 本题考查数列的求和,着重考查等差数列与等比数列的通项公式与列项相消法求和的综合应用,属于中档题.