(1)求数列{an}和{bn}的通项公式
(2)数列{cn}满足cn?anbn,求数列{cn}的前n项和Sn. 【答案】(Ⅰ)设?an?的公差为d,?bn?的公比为q 由b4?b1q3,得q3?54?27,从而q?3 2因此bn?b1?qn?1?2?3n?1
又a1?a2?a3?3a2?b2?b3?6?18?24,?a2?8 从而d?a2?a1?6,故an?a1?(n?1)?6?6n?4 (Ⅱ)cn?anbn?4?(3n?2)?3n?1
令Tn?1?30?4?31?7?32???(3n?5)?3n?2?(3n?2)?3n?1
3Tn?1?31?4?32?7?33???(3n?5)?3n?1?(3n?2)?3n
两
1式
23相
n?1减
n得
?2Tn?1?3?3?3?3?3?3???3?3
3(3n?1?1)?(3n?2)?3?1?3?3?19(3n?1?1)?（3n?2)?3n ?(3n?2)?3?1?2n73n(6n?7)?Tn??,又Sn?4Tn?7?(6n?7)?3n
4427．（ 全国大纲版高考压轴卷数学理试题）(注意:在试题卷上作答无效) ．．．．．．．．．
设数列?an?的前n项和为Sn,已知a1?8,an?1?Sn?3n?1?5,n?N?. (Ⅰ)设bn?an?2?3n,证明:数列?bn?是等比数列;
22223??(Ⅱ)证明:?a1a2a3【答案】解:(Ⅰ)
2n??1. anan?1?Sn?3n?1?5,?an?Sn?1?3n?5?n?2?,
?an?1?an?an?2?3n,即an?1?2an?2?3n?n?2?,
当n?2时,
11
nbn?1an?1?2?3n?12an?2?3n?2?3n?12?an?2?3?????2, nnnbnan?2?3an?2?3an?2?3又
b1?a1?2?31?2,b2?a2?2?32?4,?b2?2, b1?数列?bn?是以2为首项,公比为2的等比数列
(Ⅱ)由(Ⅰ)知bn?2,?an?2?3?2,?annnn?2?3n?2n,
n2n2n??an2?3n?2n22223????a1a2a31?2?????? nn2?3??3??3?2????12????2??2?232n1?2?2??2???????????an2??3?3??3?11?2?????3?n?? ??n2??2???1????n13???3????2?=??1????1.
12?3?1?328．（ 湖南省高考压轴卷数学（理）试题）数列?an?的前n项和为
Sn,a1?1,an?1?2Sn?1(n?N*),等差数列?bn?满足 b3?3,b5?9.
(1)分别求数列?an?,?bn?的通项公式; (2)设cn?bn?21(n?N*),求证cn?1?cn?.
3an?2【答案】解:(1)由an?1?2Sn?1----① 得an?2Sn?1?1----②, ①?②得an?1?an?2(Sn?Sn?1),?an?1?3an
?an?3n?1;
?b5?b3?2d?6,?d?3 ?bn?3n?6
(2)因为 an?2?3n?1,bn?2?3n
12
3nn ?3n?13n1?2n所以cn?1?cn?n?1?0
31cn?1?cn?????c1?
31所以cn?1?cn?
3
所以 cn?29．（ 天津市高考压轴卷理科数学）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) 求数列{an},{bn}的通项公式an和bn; (Ⅱ) 设cn?an?bn,求数列?cn?的前n项和Tn.
【答案】解:(Ⅰ)∵an是Sn与2的等差中项, ∴Sn?2an?2 n?2,n?N*) ② S?＝a2,Sn?2an∴? 2,S又2Sann??n－1n，（n?11?由①-②得
n? 2 an ?1, ?a? 2a n S n? 又 Sn－＝an，（n?2,n?N*) 1①
?an?2,(n?2,n?N*),即数列?an?是等比数列。 an?1
得a1?S1?2a1?2，解得a1?2。
an?0,再由Sn?2an?2∴an?2n
点（Pbn,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。(Ⅱ)
cn＝(2n?1)2n,
?anbn?1?2?3?22?5?23??(2n?1)2n,①
?Tn＝a1b1?a2b2??2Tn?1?22?3?23??(2n?3)2n?(2n?1)2n?1. ②
23①-②得:?Tn?1?2＋(2?2＋2?2＋即:?Tn?1?2?(2?2?∴Tn?(2n?3)2
n?134＋2?2n)?(2n?1)2n?1,
?2n?1）?(2n?1)2n?1,
?6
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30．（ 陕西省高考压轴卷数学（理）试题）在等比数列{an}中,已知a1?3,公比q?1,等
差数列{bn}满足b1?a1，b4?a2，b13?a3. (Ⅰ)求数列{an}与{bn}的通项公式;
(Ⅱ)记cn?(?1)nbn?an,求数列{cn}的前n项和Sn.
【答案】【解析】(Ⅰ) 设等比数列?an?的公比为q,等差数列?bn?的公差为d. 由已知得:a2?3q,a3?3q2,
b1?3,b4?3?2d,b13?3?12d
?3q?3?3d?q?1?d??q?3或 q?1(舍去), ?2?2?3q?3?12d?q?1?4d所以, 此时 d?2
所以,an?3n, bn?2n?1. (Ⅱ) 由题意得:cn?(?1)nbn?an?(?1)n(2n?1)?3n
Sn?c1?c2???cn
?(?3?5)?(?7?9)???(?1)n?1(2n?1)?(?1)n(2n?1)?3?32???3n 3n?133n?13当n为偶数时,Sn?n????n?
22223n?133n?17当n为奇数时,Sn?(n?1)?(2n?1)????n?
2222?3n?13?n???22所以,Sn??n?1?3?n?7?2?2(n为偶数时) .
(n为奇数时)1*
,其中n?N. 4an31．（ 福建省高考压轴卷数学理试题）已知数列?an?满足a1?1,an?1?1?(Ⅰ)设bn?,求证:数列?bn?是等差数列,并求出?an?的通项公式an;
2an?14an(Ⅱ)设cn?,数列?cncn?2?的前n项和为Tn,是否存在正整数m,使得
n?12 14
Tn?1*
对于n?N恒成立,若存在,求出m的最小值,若不存在,请说明理由.
cmcm?1【答案】【解析】(I)证明
bn?1?bn?22an?1?12an?1?2?22an?1?1???2?1???14an???2?4an2??2,
2an?12an?1所以数列?bn?是等差数列,a1?1,b1?2,因此bn?2?(n?1)?2?2n,由
bn?22an?1得an?n?1. 2n,
(II)
cn?2ncncn?2?4]??1?2???n?n?2??nn?2?,所以
11??1Tn?2?1?????3,
2n?1n?2??依题意要使Tn?m(m?1)1对于n?N*恒成立,只需?3,
4cmcm?1解得m?3或m??4,所以m的最小值为3.
32．（ 广东省高考压轴卷数学理试题）设数列?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)设bn?{}an,求证:b1?b2?...?bn?1. 2Sn【答案】证明:(Ⅰ)?Sn+1=3Sn+2,∴Sn+1+1=3(Sn+1), 又?S1+1=3,
∴{Sn+1}是首项为3,公比为3的等比数列,且Sn?3n?1,n?N*
(Ⅱ)当n=1时,a1=S1=2,
当n?2时,an?Sn?Sn?1?(3n?1)?(3n?1?1) ?3故an?2?3n?1,n?N*
n?1(3?1) ?2?3n?1.
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