数列?an?满足an?d1?d2?d3?????d2n;数列?bn?为公比大于1的等比数列,且b2,b4为方程
x2?20x?64?0的两个不相等的实根.
(Ⅰ)求数列?an?和数列?bn?的通项公式;
(Ⅱ)将数列?bn?中的第．a1项,第．a2项,第．a3项,,第．an项,删去后剩余的项按从小到大的顺序排成新数列?cn?,求数列?cn?的前2013项和.
3?(?1)n【答案】解:(Ⅰ)?dn? ,
2?an?d1?d2?d3?????d2n?23?2n?3n 2因为b2,b4为方程x?20x?64?0的两个不相等的实数根. 所以b2?b4?20,b2?b4?64 解得:b2?4,b4?16,所以:bn?2n
(Ⅱ)由题知将数列?bn?中的第3项、第6项、第9项删去后构成的新数列?cn?中的奇数列与偶数列仍成等比数列,首项分别是b1?2,b2?4公比均是8,
T2013?(c1?c3?c5?????c2013)?(c2?c4?c6?????c2012)
2?(1?81007)4?(1?81006)20?81006?6???
1?81?87
40．（山东省凤城高中2013届高三4月模拟检测数学文试题 ）已知数列?an?是等差数列,?bn?是等比数
列,且a1?b1?2,b4?54,a1?a2?a3?b2?b3. (Ⅰ)求数列?an?和?bn?的通项公式
(Ⅱ)数列?cn?满足cn?anbn,求数列?cn?的前n项和Sn. 【答案】解:(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
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(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?1n减得
?2Tn?1?3?3?3?3?3?3???3?33(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
4441．（山东省聊城市2013届高三高考模拟（一）文科数学）已知正项数列?an?的前n项和为Sn,且
a1?1,an?Sn?Sn?1(n?2)
(I)求数列?an?的通项公式;
2an?3(Ⅱ)设bn?2?1,数列?bn?的前项n和为Tn,求证:Tn?n?1【答案】
an?1?1
42．（山东省德州市2013届高三3月模拟检测文科数学）
数列{an}是公差不小0的等差数列a1、a3,是函数f(x)?1n(x?6x?6)的零点,数列{bn}的前n项和为Tn,且Tn?1?2bn(n?N*) (1)求数列{an},{bn}的通项公式;
(2)记cn?anbn,求数列{cn}的前n项和Sn. 【答案】
2 17
43．（山东省莱芜市莱芜十七中2013届高三4月模拟数学（文）试题）已知数列{an}的前n项和Sn满足
Sn?2an?1,等差数列{bn}满足b1?a1,b4?S3.
(1)求数列{an}、{bn}的通项公式; (2)设cn?10011,数列{cn}的前n项和为Tn,问Tn>的最小正整数n是多少?
2012bnbn?1【答案】解:(1)当n?1时,a1?S1?2a1?1,∴a1?1
当n?2时,an?Sn?Sn?1?(2an?1)?(2an?1?1)?2an?2an?1, 即
an?2 an?1∴数列{an}是以a1?1为首项,2为公比的等比数列,∴an?2n?1,Sn?2n?1 设{bn}的公差为d,b1?a1?1,b4?1?3d?7,∴d?2 ∴bn?1?(n?1)?2?2n?1 (2)cn?11111??(?) bnbn?1(2n?1)(2n?1)22n?12n?118
11111111n(1????...??)?(1?)? 23352n?12n?122n?12n?11001n1001由Tn>,得>,解得n>100.1
20122n?120121001∴Tn>的最小正整数n是101
2012∴Tn?44．（山东省泰安市2013届高三第二次模拟考试数学（文）试题 ）已知等差数列?an?的首项
a1?3,公差d?0,其前n项和为Sn,且a1,a4,a13分别是等比数列?bn?的b2,b3,b4.
(I)求数列?an?与?bn?的通项公式; (II)证明13?1S?1?????1?3. 1S2Sn4【答案】
19
45．（山东省日照市2013届高三第一次模拟考试数学（文）试题）若数列?bn?:对于n?N,都有
?bn?2?bn?d(常数),则称数列?bn?是公差为d的准等差数列.如数列cn:若
?4n?1,当n为奇数时;cn??则数列?cn?是公差为8的准等差数列.设数列?an?满足:a1?a,对于
?4n?9,当n为偶数时.n?N?,都有an?an?1?2n.
(I)求证:?an?为准等差数列;
(II)求证:?an?的通项公式及前20项和S20. 【答案】解:(Ⅰ)?an?an?1?2n(n?N)① ∴an?1?an?2?2(n?1) ② ②-①,得an?2?an?2(n?N). 所以,?an?为公差为2的准等差数列
(Ⅱ)又已知a1?a,an?an?1?2n(n?N),∴a1?a2?2?1,即a2?2?a. 所以,由(Ⅰ)a1,a3,a5,?成以a为首项,2为公差的等差数列,
???a2,a4,a6,?成以2?a为首项,2为公差的等差数列,所以
当n为偶数时,an?2?a???n??1??2?n?a, 2??当n为奇数时,an?a???n?1??1??2?n?a?1.
?2??n?a?1,n为奇数, ?an???n?a, n为偶数.S20a S??a20n?a1?a2?a3?a4???an?119?(a1?a2)?(a3?a4)???((a19??aan?120n)
?2?1?2?3???2?19 (n?1=2?(1?19)?10?200 246．（山东省济宁市2013届高三第一次模拟考试数学（文）试题 Word版含答案）(本小题满分l2分)
设数列{an}满足:a1=5,an+1+4an=5,(n?N*)
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