2013届高三数学(文)专题强化测评(十二)

专题强化测评（十二）一、选择题1.(2011·莆田模拟)已知等差数列{an}满足a2+a8=16,则a5等于()(A)10(B)8(C)6(D)42.已知各项不为0的等差数列{an}满足2a3-a72+2a11=0，数列{bn}是等比数列，且b7=a7,则b5b9的值为()(A)2(B)4(C)8(D)163.(2011·聊城模拟)已知各项不为0的等差数列{an},满足2a3-a72+2a11=0,数列{bn}是等比数列，且b7=a7,则b6b8=()(A)2(B)4(C)8(D)164.已知数列{an}满足log3an+1=log3an+1(n∈N*)且a2+a4+a6=9,则15793logaaa的值是()(A)-5(B)-15(C)5(D)15二、填空题5.已知在正项等比数列{an}中,a1=1,a3a7=4a62,则S6=_____.6.设Sn为等比数列{an}的前n项和,8a2-a5=0,则42SS=_____.7.(2011·枣庄)设Sn是等差数列{an}的前n项和，S5=3(a2+a8),则53aa的值为_____.三、解答题8.(2011·北京模拟)已知数列{an}的前n项和为Sn,且Sn=4an-3(n∈N*)．(1)证明:数列{an}是等比数列；(2)若数列{bn}满足bn+1=an+bn(n∈N*)，且b1=2，求数列{bn}的通项公式．9.(2011·威海模拟)已知等差数列{an}的前n项和为Sn，公差d≠0,且S3+S5=50,a1,a4，a13成等比数列．(1)求数列{an}的通项公式；(2)设{nnba}是首项为1，公比为3的等比数列，求数列{bn}的前n项和Tn.10.(2011·泉州模拟)已知数列{an}是各项均不为0的等差数列，Sn为其前n项和，且满足an2=S2n-1,n∈N*.(1)求数列{an}的通项公式；(2)数列{bn}满足bn=nn11,aa求证：数列{bn}的前n项和Tn＜1.211.(2011·嘉兴模拟)已知数列{an}满足a1=3,an+1-3an=3n(n∈N*)，数列{bn}满足bn=3-nan.(1)求证：数列{bn}是等差数列；(2)设Sn=312naaaa345n2，求满足不等式n2nS11128S4的所有正整数n的值.专题强化测评（十二）1.【解析】选B.a2+a8=2a5=16,∴a5=8.2.【解】选D.由2a3-a72+2a11=0得a72-4a7=0，又a7≠0,∴a7=4，∴b7=a7=4，b5b9=b72=16.3.【解析】选D.由题意得a72=2(a3+a11)=4a7,∴a7=4,b7=4,∴b6·b8=b72=16.4.【解析】选A.由log3an+1=log3an+1(n∈N*)，得an+1=3an,所以数列{an}是公比为3的等比数列，因为a2+a4+a6=9,所以a5+a7+a9=(a2+a4+a6)×33=35,所以15793logaaa=-log335＝-5.5.【解析】由题意知a3a7=a52=4a62，∴a5=2a6,∴6665a11q63q.S.a21q326.【解析】由8a2-a5=0,得q3=8,q=2,∴42422S1q1q5.S1q7.【解析】15aa2×5=3×2a5,5a3=6a5,∴53aa=5.6答案：568.【解析】(1)由Sn=4an-3,当n=1时，a1=4a1-3,解得a1=1.因为Sn＝4an-3,则Sn-1=4an-1-3(n≥2),所以当n≥2时，an=Sn-Sn-1=4an-4an-1,整理得nn14aa3,又a1=1≠0,所以{an}是首项为1，公比为43的等比数列.(2)因为an=(43)n-1,由bn+1=an+bn(n∈N*),得bn+1-bn=(43)n-1.可得bn=b1+(b2-b1)+(b3-b2)+…+(bn-bn-1)=n1n141()4323()14313(n≥2,n∈N*).当n=1时上式也满足条件.所以数列{bn}的通项公式为bn=3×(43)n-1-1(n∈N*).9.【解析】(1)依题意得11211132453ad5ad5022a3daa12d,解得1a3d2,∴an=a1+(n-1)d=3+2(n-1)=2n+1,即an=2n+1(n∈N*)(2)n1n1n1nnnnb3,ba32n13,aTn=3+5·3+7·32+…+(2n+1)·3n-1,3Tn=3·3+5·32+7·33+…+(2n-1)·3n-1+(2n+1)·3n,两式相减得-2Tn=3+2·3+2·32+…+2·3n-1-(2n+1)3n=n1nn313322n132n3,13∴Tn=n·3n(n∈N*)10.【解析】(1)(方法一)设等差数列{an}的公差为d,首项为a1,在an2=S2n-1中，令n=1,n=2,221111222311aaaS,,aSad3a3d得即解得a1=1,d=2,∴an=2n-1.(方法二)∵{an}是等差数列，12n1naaa2，∴S2n-1=12n1aa2(2n-1)=(2n-1)an.由an2=S2n-1,得an2=(2n-1)an,又∵an≠0,∴an=2n-1,则a1=1,d=2.∴an=2n-1.nnn1nn112baa2n12n1111(),22n12n1111111nT(1).23352n12n12n11n2n12n10,22n122n122n11T.2＞＜11.【解析】(1)由bn=3-nan得an=3nbn，则an+1=3n+1bn+1.代入an+1-3an=3n中，得3n+1bn+1-3n+1bn=3n，即得bn+1-bn=13.所以数列{bn}是以1为首项，以13为公差的等差数列.(2)因为数列{bn}是首项为b1=3-1a1=1，公差为13的等差数列，则n1n2b1n133，则nn1nna3bn23.从而有n1na3n2，故312nnaaaaS345n2=1+3+32+…+3n-1=nn1331,132则nn2nn2nS311S3131，由n2nS11128S4，得n111.128314即33n127，得1n≤4.故满足不等式n2nS11128S4的所有正整数n的值为2，3，4.