?数列{cn}是以?2d为公差的等差数列3
2(2)
?a1?a3???a25?130
a2?a4???a26?143?13k
?两式相减:13d?13?13k
?d?1?k
13(13?1)?13a1??2d?1302
a1??2?12k
?an?a1?(n?1)d?(1?k)n?(13k?3)
22?cn?an?an?1?(an?an?1)(an?an?1)
?26k2?32k?6?(2n?1)(1?k)2 ??2(1?k)2n?25k2?30k?58
S(3)因为当且仅当n?12时n最大 ?有c12?0,c13?0
22????24(1?k)?25k?30k?5?0?k?18k?19?0??2?22??36(1?k)?25k?30k?5?0??k?22k?21?0 即??k?1或k??19???k??19或k?21?k?21或k?112
33．（山东省曲阜师大附中2013届高三4月月考数学（文）试题）设数列{an}的前n项和为Sn,且满足
Sn?1?2an,n?N*.
(1)求数列{an}的通项公式;
(2)在数列{an}的每两项之间都按照如下规则插入一些数后,构成新数列:an和an?1两项之间插入n个数,使这n?2个数构成等差数列,其公差记为dn,求数列?【答案】
?1??的前n项的和Tn. d?n? 11
34．（山东省莱芜五中2013届高三4月模拟数学（文）试题）在等差数列{an}中,a3?a4?a5?42,a8?30.
(1)求数列{an}的通项公式; (2)若数列?bn?满足bn?(3)an?2??(??R),则是否存在这样的实数?使得?bn?为等比数列;
?2n?1,n为奇数?(3)数列?cn?满足cn??1，Tn为数列?cn?的前n项和,求T2n.
?an?1,n为偶数?2【答案】解:(1)因为{an}是一个等差数列,所以a3?a4?a5?3a4?42,?a4?14. 设数列{an}的公差为d,则4d?a8?a4?16,故d?4;故an?a4?(n?4)d?4n?2 (2)bn?(3)an?2???9n??.
假设存在这样的?使得?bn?为等比数列,则bn?12?bn?bn?2,即(9n?1??)2?(9n??)?(9n?2??), 整理可得??0. 即存在??0使得?bn?为等比数列
?2n?1,n为奇数(3)∵cn??,
?2n?3,n为偶数∴T2n?1?(2?2?3)?2?(2?4?3)?2???2242n?2?(2?2n?3)
?1?22?24???22n?2?4(1?2???n)?3n
1?4nn(n?1)4n?1??4??3n??2n2?n 1?423
12
35．（山东省济南市2013届高三3月高考模拟文科数学）正项等比数列{an}的前n项和为Sn,a4?16,
且a2,a3的等差中项为S2. (1)求数列{an}的通项公式; (2)设bn?na2n?1,求数列{bn}的前n项和Tn .
【答案】解:(1)设等比数列{an}的公比为q(q?0),
3??a1?2?a1q?16由题意,得?,解得 ?2??q?2?a1q?a1q?2(a1?a1q)所以an?2n (2)因为bn?所以Tn?na2n?1?n22n?1,
1234n?3?5?7???2n?1, 222221123n?1nTn???????, 423252722n?122n?1311111n所以Tn??3?5?7???2n?1?2n?1
422222211(1?n)4?n?2?4?3n ?22n?1133?22n?121?4816?12n故Tn??
99?22n?136．（山东省滨州市2013届高三第一次（3月）模拟考试数学（文）试题）已知数列?an?的前n项和是Sn,
且Sn?1an?1(n?N?) 2(Ⅰ)求数列?an?的通项公式;
(Ⅱ)设bn?log1(1?Sn?1)(n?N),令Tn?3?111???,求Tn. b1b2b2b3bnbn?1【答案】
13
37．（山东省莱钢高中2013届高三4月模拟检测数学文试题 ）设数列?an?为等差数列,且
a5?14,a7?20,数列?bn?的前n项和为Sn,b1?(Ⅰ)求数列?an?,?bn?的通项公式;
2且3Sn?Sn?1?2(n?2,n?N);, 3(Ⅱ)若cn?an?bn,n?1,2,3,?,Tn为数列?cn?的前n项和. Tn ?【答案】解:(Ⅰ) 数列?an?为等差数列,公差d?易得a1?2 所以 an?3n?1 1(a7－a5)?3 , 2由bn?2－2Sn,令n?1,则b1?2?2S1,又S1?b1,所以. b2?2?2(b1?b2),则b2?由3Sn?Sn?1?22 9当n?3时,得3Sn?1?Sn?2?2, 两式相减得.3(Sn?Sn?1)?Sn?1?Sn?2即3bn?bn?1 .所以?bn?是以b1?bnb11? 又2? b13bn?1321为首项,为公比的等比数列, 33 14 于是bn?2?1 3n1 n31111∴Tn?2[2??5?2?8?3???(3n?1)?n], 3333(Ⅱ)cn?an?bn?2(3n?1)?1111??1Tn?2?2?2?5?3???(3n?4)?n?(3n?1)?n?1? 3333??32111111?3?2?3?3???3?n??(3n?1)?n?1] 3333333771n所以 Tn???n?n?1 2233771n7从而Tn???n?n?1? 2232377?∵T n 22两式相减得Tn?2[3?38．（山东省烟台市2013届高三3月诊断性测试数学文）设{an}是正数组成的数列,a1=3.若点 ?an,an?12?2an?1?(n?N*)在函数f(x)?(1)求数列{an}的通项公式; (2)设bn?【答案】 13x?x2?2的导函数y?f?(x)图像上. 32*,是否存在最小的正数M,使得对任意n?N都有b1+b2++bn an?1?an 3?(?1)n39．（山东省青岛市2013届高三第一次模拟考试文科数学）已知n?N,数列?dn?满足dn?, 2? 15