在Tn?1?12bn中,令n?1,得b1?121223.当n?2时,Tn=1?1312bn,Tn?1?1?12bn?1,
两式相减得bn?bn?1?bn,?bnbn?1??n?2?
2?1??bn???3?3?(2)
n?1?23n23n?n?N?
?cn??2n?1???4n?23n,
352n?1?Sn32n?32n?1??1?1,?Sn?2??2?3????2????????, n23nn?133333?33?3?3???1?1?2?1?????n?112n?19?3??=2???n?113?3?1???3????111?2n?1??1Sn?2??2?2?3???n??n?1?333?3?3?3?2
=
12n?1?44n?4?112???n????n?1n?133?333?32n?23n,
?Sn?2?10．已知数列
?an?的前n项和为Sn,且an是Sn与2的等差中项,数列?bn?中,b1?1,点P?bn,bn?1? 在直线x?y?2?0上?
(Ⅰ)求
a1和a2的值; (Ⅱ)求数列?an?,?bn?的通项an和bn; (Ⅲ) 设cn?anbn,求数列?cn?的前n项和Tn?
an是Sn与2的等差中项, ∴Sn【解析】(1)∵
?2an?2?
∴a1(2)??S1?2a1?2解得a1?2, a1?a2?S2?2a2?2 解得a2?4
Sn?2an?2 Sn?1?2an?1?2 又Sn?Sn?1?ann?2,n?N???
?an?2an?2an?1 又?an?0
anan?1??2,n?2,n?N??? 即数列?a?是等比数列
n?a1?2 ?an?2n 又?点P?bn,bn?1?在直线x?y?2?0上,
?bn?bn?1?2?0 ?bn?1?bn?2,即数列?bn?是等差数列,又b1?1,?bn?2n?1
(3)
?cn＝(2n?1)2,n
23n?Tn＝a1b1?a2b2???anbn?1?2?3?2?5?2???(2n?1)2,第 6 页 共 48 页
?2Tn?1?2?3?2???(2n?3)2?(2n?1)223nn?1?
因此由错位相减法得,∴
Tn?(2n?3)2n?1?6?
11．已知在等差数列
?an?中,a3?4,前7项和等于35,数列?bn?中,点?bn,Sn?在直线
x?2y?2?0上,其中Sn是数列
?bn?的前n项和
?n?N??
*(1)求数列
?an?的通项公式; (2)求证数列?bn?是等比数列;
并证明;
(3)设
cn?an?bn,Tn为数列?cn?的前n项和,求Tn43?Tn?52?
?a1?2d?4?a1?2?【解析】(1)设数列{an}的公差为d,则由题意知?得? 7?6d?17a?d?35??1?2∴an?a1?(n?1)d?2?n?1?n?1.
(2)∵点
(bn,Sn)在直线x?2y?2?0上
∴bn?2Sn?2?0----① ,bn?1?2Sn?1?2?0 (n?2) -----②
①-②得bn又当n?bn?1?2bn?012,∴bn?13bn?1(n?2),
23?0
?1时,b1??2323b1?1 ∴b1?13n?1∴数列{bn}是以
为首项,为公比的等比数列?
(3)由(2)知,
bn??(?13)?223nn,
∴cn32?22?32?42(n?1)Tn??????23n333313Tn?2?22?an?bn?(n?1)?
-----------③
③—④得,
3333322?22222(n?1)Tn??2?3??n?n?133333313132?2?33?2?44???2nn?2(n?1)n?1------④
1∴Tn?2???133???13n?1?(n?1)3n3=2?(1?1?131n?1)?n?13n
3=
2?12(1?13)?n?1n?153n=
2?2n?52?3n
第 7 页 共 48 页
Tn?52?2n?52?3n?52
由③知Tn的最小值是T112．设数列
?43 ∴
43?Tn?52
?an?的前n项和为Sn,且满足S1=2,Sn+1=3Sn+2?n=1,2,3??.
(Ⅰ)求证数列
{Sn+1}为等比数列; (Ⅱ)求通项公式an; (Ⅲ)设bn?anS2n,求证b1?b2?...?bn?1.
【解析】证明(Ⅰ)?Sn+1=3Sn+2, ∴Sn+1+1=3(Sn+1).
又?S1+1=3,
n*∴{Sn+1}是首项为3,公比为3的等比数列且Sn?3?1,n?N(Ⅱ)
.
n=1时,a1=S1=2,
nn?1n>1时,an?Sn?Sn?1?(3?1)?(3?1) ?3n?1(3?1) ?2?3n?1. 故an?2?3n?1,n?N*.
(Ⅲ)
?bn?2?3nn?12(3?1)?2?3(312n?1n?1n?1)(3?1)1?1?13n?1?11?13?1?n,?n?1?
113n?1?b1?b2?...?bn??12?12?13?1n?(3?113?12)?(3?123?13)?????(?1?13?1n)
?1.
【命题意图】 数列既是高中数学的重点,也是难点.掌握好等差、等比数列的通项公式和前n项和公式,能用概念判断是否为等差、等比数列.常见考点
Sn与an的关系(注意讨论);
an?1?kan?b;递推——猜想——数学归纳法证明;迭加
an?1?an?f(n);迭乘an?1?f(n)?an;
裂项求和;错位相减等;数列不等式证明中注意放缩法的运用. 13．已知等差数列{an}的首项a1?2,公差d?0,且第一项、第三项、第十一项分别是等比数列{bn}的第一项、第二项、第三项?
(I)求数列{an}和{bn}的通项公式;
?(II)设数列{cn}对任意的n?N均有c1b1?c2b2???cnbn?an?1,求数列{cn}的前n项和?
【解析】(I)由已知
(2?2d)?2(2?10d)?d?32ord?0(舍)
数列{an}的通项公式an?3n?1;数列{bn}的通项公式bn?2h2n?1
(II)由
c1b1?c2b2???cnbn?an?1,c1b1?c2b2???cn?1bn?1), ?an?cnbn?an?1?an?3(n?2)
cn?3?22n?1(n?2)
又c1?b1?a2?10
?10,(n?1)cn??
2n?13?2,(n?2)?第 8 页 共 48 页
所以数列{cn}的前n项和S?10?24(1?4n1?414．设数列{an}的首项a1n?1)?2?22n?1
?1,前n项和为Sn,且点(Sn?1,Sn)(n?N,n?2)在直线(2t?3)x?3ty?3t?0(t?为与n无关的正实
数)上.
(Ⅰ) 求证数列{an}是等比数列;
(Ⅱ) 记数列{an}的公比为f(t),数列{bn}满足b1?1,b?n?f(1b)(n?N,n?2).
n?1设cn?b2n?1b2n?b2nb2n?1,求数列{cn}的前n项和Tn;
(Ⅲ) 在(Ⅱ)的条件下,设dn*n?(1?13b)(n?N),证明dn?dn?1.
n?1【解析】(Ⅰ)因为点
(Sn?1,S?n)(n?N,n?2)在直线
(2t?3)x?3ty?3t?0(
t为与
n无关的正实数)上, (2t?3)Sn?1?tS3n?t3?,0即有3tS?(2t?3)S?nn?1?3t(n?N,n?2).
当n?2时,3t(a1?a2)?(2t?3)a1?3t.
由a31?1,解得a2?2t?33t,所以
a2a?2t?13t.
当n?2时，有
3tSn＋1?(2t?3)Sn?3t ① 3tSn?(2t?3)Sn?1?3t ②
①-②,得
3ta3n＋1?(2t?3)aan?1n?0,整理得
a?2t?n3t.
综上所述,知
an?1?3a?2t
n3t
(n?N*),因此{an}是等比数列 1?3(Ⅱ) 由(Ⅰ) 知
f(t)?2t?32?3t,从而bn?f(1b)?bn?1n?11?23?bn?1,
3?bn?1所以b?n?bn?1?23(n?N,n?2).
因此,{bn}是等差数列,并且bn?b1?(n?1)d?23n?13.
所以,Tn?c1?c2?c3???cn
第 9 页 共 48 页
所以
?b1b2?b2b3?b3b4?b4b5???b2n?1b2n?b2nb2n?1 ?b2(b1?b3)?b4(b3?b5)???b2n(b2n?1?b2n?1)
4n(b2?b2n)4??2d(b2?b4???b2n)??????323824??n?n
93(Ⅲ) 由(Ⅱ)知dnn(53?4n?132)
?(1?12n)n,则dn?1??1??1??2n?1????n?1.
1??将d?1?n??2n??knn用二项式定理展开,共有n?1项,其第k?1项?0?k?n?为
Tk?112k11n?n?1???n?k?1??1? ?C?????kk2k!n?2n??1?1??2??k?1???1???1????1??, k!?n??n??n? n?1用二项式定理展开,共有
k???1同理,dn?1??1??2n?1????项中的第kn?2项,第n?2项为Un?2?Cn?1n?1??1??2n?1????n?1?0,其前n?1?1项?0?k?n?为Uk?1?1n?12n2n?112k?1?1??2??k?1???1?1??1??????, k!?n?1??n?1??n?1?kn?1?kn?1,k?2,3,?,n,
,
由1?1n?1?,1??1?,?,1?得Tk?1?Uk?1,k?2,3,?,n,又T1?U1,T2?U2,Un?2?0∴
dn?dn?1
15．已知递增的等比数列
(Ⅰ)求数列
?an?满足a2?a3?a4?28,且a3?2是a2,a4的等差中项.
?an?的通项公式;
2(Ⅱ)若
bn?logan?1,Sn是数列
?bn?的前n项和,求使Sn?42?4n成立的n的最小值.
【解析】(Ⅰ)设等比数列
?an?的公比为q,依题意有2(a3,将(1)代入得a3?2)?a2?a4, (1)
.
又a2?a3?a4?28?8.所以a2?a4?20?a1q?a1q3?20,于是有?
2a1q?8,?第 10 页 共 48 页