?1?q?2 q21?q?1或q??
233当q?1时,a1?,an?
221n?11当q??时,a1?6,an?6(?)
22所以
或解:当q?1时由条件得:
3?2aq???12?a(1?q3)9 ?1??2?1?q1?q332,即2q?3q?1?0 ?32q(1?q)1?(2q?1)(1?q)2?0 ?q??
2?a1?6
当q?1时,a1?所以
(2)当q?1时,Sn?3符合条件 23(1?2???n) 23n(n?1) 4101121n?11当q??时,Sn?6[(?)?2?(?)?3?(?)???n(?)]
2222211111??Sn?6[(?)?2?(?)2?3?(?)3???n(?)n]
2222231111?Sn?6[1?(?)?(?)2???(?)n?1?n(?)n] 2222211?(?)n32?n(?1)n] ?Sn?6[1221?2841?Sn??(3n?2)(?)n
332?[来源:学+科+网Z+X+X+K]22．（广东省海珠区2013届高三上学期综合测试（一）数学（理）试题）(本小题满分14分)
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已知等差数列?an?满足a3?5,a5?2a2?3,又数列?bn?中,b1?3且
3bn?bn?1?0?n?N??.
(1)求数列?an?,?bn?的通项公式;
(2)若数列?an?,?bn?的前n项和分别是Sn,Tn,且cn?(3)若Mn?9logmSn?2Tn?3?.求数列?cn?的前n项和Mn;
n3?m?0,且m?1?对一切正整数n恒成立,求实数m的取值范围. 4【答案】(本小题主要考查数列通项、错位求和与不等式等知识,考查化归与转化、方程的数学思想方
法,以及运算求解能力)
解: ( 1)设等差数列?an?的公差为d,则由题设得:
??a1?2d?5 ???a1?4d?2?a1?d??3即??a1?2d?5?a1?1,解得?
?a?2d?3d?2??1[来源:学科网]?an?1??n?1??2?2n?1?n?N??
3bn?bn?1?0
?bn?1?3,?n?N?? bn?数列?bn?是以b1?3为首项,公比为3的等比数列 ?bn?3?3n?1?3n?n?N??.
n?1?2n?1??n2, (2)由(1)可得Sn?23?3n?31n?1Tn???3?3?.
1?32?cn?n2?3n?1?3?3?n?n?3n?1. ?cn?1?cn
?Mn?c1?c2?c3?Mn?1?32?2?33?3?34???n?1??3n?n?3n?1?1?
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3Mn?1?33?2?34?3?35???n?1??3n?1?n?3n?2?3n?1?n?3n?2
?2?
?1???2?得: ?2Mn?32?33?34?32?3n?1?3??n?3n?2
1?3?Mn9n???2n?1?3?1n?N????.?4?99n?1??2n?1?3?1?(3)Mn?1?Mn?????2n?1??3n?1????? 44?
?9?n?1??3n?0
Mn?1?Mn,?n?N??
?当n?1时, ?Mn取最小值,M1?9, ?9?9logm即logm3 43?1 43当m?1时,logm?1恒成立;
43当0?m?1时,由logm?1 ?logmm,
43得m? ,
43?0?m?.
43???实数m的取值范围是?m0?m?或m?1?
4??23．（广东省惠州市2013届高三一调(理数)）等差数列{an}中,a1?1,前n项和为Sn,等比数列{bn}各项均
为正数,b1?2,且s2?b2?7,s4?b3?2. (1)求an与bn; (2)设cn?a2n?11?, Tn?c1?c2?c3???cn 求证:Tn? (n?N). a2n2n【答案】解:(1)设等差数列
?an?的公差为d,等比数列{bn}的公比为q,
2由题知: s2?b2?7, s4?b3?2 ?d?2q?5,3d?q?1?0
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解直得,q=2或q=-8(舍去),d=1;
?an?1?(n?1)?n bn?2n;
(2)证明:?cn?a2n?12n?11352n?1,?cn? .Tn??????
2n2462na2n12n对一切正整数成立.
法一、 下面用数学归纳法证明Tn?(1)当n?1时，T1?12?1?1?,命题成立 22?1?Tk?(2)假设当n?k时命题成立，12k
?Tk?1?Tk?则当n?k?1时，2k?112k?112k?1?=
2(k?1)2k2(k?1)2k?12kk?114k2?4k?1=,这就是说当n?k?1时命题成立 ?24k?4k2k?12k?11综上所述原命题成立 法二、??Tn2n?1n? n?2n?11133552n?12n?11123452n?22n?11????????????????? 2244662n2n2234562n?12n4n?Tn?12n
[来源:学.科.网Z.X.X.K]
法三、设数列?An?,An?nTn,则An?1?n?1Tn?1
4n2?4n?1?1 24n?4n1 2A(2n?1)n?12n?1?n?1???An2(n?1)n2n?1n?数列?An?单调递增,于是An?An?1???A1,而A1??Tn?12n
24．（2013届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;
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{} (Ⅱ)设bn?an,求证:b1?b2?...?bn?1. 2Sn=3Sn+2,∴Sn+1+1=3(Sn+1),
【答案】证明:(Ⅰ)?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.
2?3n?12?3n?111bn?n???,?n?2?
(3?1)2(3n?1?1)(3n?1)3n?1?13n?1?b1?b2?...?bn??1111111?(1?2)?(2?3)?????(n?1?n) 23?13?13?13?13?13?1111??n?1 223?125．（广东省汕头市2013届高三上学期期末统一质量检测数学（理）试题）已知有两个数列{an},{bn},它们
的前n项和分别记为Sn,Tn,且数列{an}是各项均为正数的等比数列,Sm=26,前m项中数值最大的项的值,18,S2m=728,又Tn?2n2 (I)求数列{an},{bn}的通项公式.
(II)若数列{cn}满足cn?bnan,求数列{cn}的前n项和Pn.
【答案】解:(Ⅰ)设等比数列
?an?的公比为q ,
an?0 , ?q?0
若q=1时 Sm?ma1 S2m?2ma1 此时2Sm?S2m 而已知 Sm?26 S2m?728
?2Sm?S2m , ?q?1
?a1?1?qm???26?1?S?26?m?1?q由? 得 ?
2mS?728?m?a1?1?q??728?2???1?q无限精彩在大家 www.TopSage.com