在等差数列{an}中,a1=3,其前n项和为Sn,等比数列{bn}的各项均为正数,b1=1,公比为q,且b2+S2=12.q=S2b2 (Ⅰ)求an与bn; (Ⅱ)设数列{cn}满足cn=1/Sn,求的{cn}的前n项和Tn.

问题描述:

在等差数列{an}中,a1=3,其前n项和为Sn,等比数列{bn}的各项均为正数,b1=1,公比为q,且b2+S2=12.q=

S2
b2

(Ⅰ)求an与bn
(Ⅱ)设数列{cn}满足cn=
1
Sn
,求的{cn}的前n项和Tn

(Ⅰ)设{an}的公差为d,
因为

b2+S2=12
q=
S2
b2

所以b2+b2q=12,即q+q2=12,
∴q=3或q=-4(舍),
b2=3,s2=9,a2=6,d=3.
故an=3+3(n-1)=3n,bn=3n-1
(Ⅱ)因为Sn=
n(3+3n)
2
=
3n(n+1)
2

所以:cn=
1
Sn
=
2
n(3+3n)
=
2
3
(
1
n
-
1
n+1
)

故Tn=
2
3
[(1-
1
2
)+(
1
2
-
1
3
)+…+(
1
n
-
1
n+1
)]=
2
3
(1-
1
n+1
)=
2n
3(n+1)