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=
,求的{cn}的前n项和Tn. 1 Sn
答
(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=
[(1-2 3
)+(1 2
-1 2
)+…+(1 3
-1 n
)]=1 n+1
(1-2 3
)=1 n+1
.2n 3(n+1)