百度文库,精选习题试题习题,尽在百度训练目标(1)数列知识的综合应用;(2)中档大题的规范练.训练题型(1)等差、等比数列的综合;(2)数列与不等式的综合;(3)数列与函数的综合;(4)一般数列的通项与求和.解题策略(1)将一般数列转化为等差或等比数列;(2)用方程(组)思想解决等差、等比数列的综合问题.1.设数列{an}的前n项和为Sn.已知2Sn=3n+3.(1)求{an}的通项公式;(2)若数列{bn}满足anbn=log3an,求{bn}的前n项和Tn.2.已知数列{an}是递增的等比数列,且a1+a4=9,a2a3=8.(1)求数列{an}的通项公式;(2)设Sn为数列{an}的前n项和,bn=an+1SnSn+1,求数列{bn}的前n项和Tn.百度文库,精选习题试题习题,尽在百度3.已知数列{an}的各项均为正数,Sn是数列{an}的前n项和,且4Sn=a2n+2an-3.(1)求数列{an}的通项公式;(2)已知bn=2n,求Tn=a1b1+a2b2+⋯+anbn的值.4.在数列{an}中,a1=12,其前n项和为Sn,且Sn=an+1-12(n∈N*).(1)求an,Sn;(2)设bn=log2(2Sn+1)-2,数列{cn}满足cn·bn+3·bn+4=1+(n+1)(n+2)·2bn,数列{cn}的前n项和为Tn,求使4Tn>2n+1-1504成立的最小正整数n的值.5.已知函数f(x)满足f(x+y)=f(x)·f(y)且f(1)=12.(1)当n∈N*时,求f(n)的表达式;(2)设an=n·f(n),n∈N*,求证:a1+a2+a3+⋯+an<2;(3)设bn=(9-n)fn+1fn,n∈N*,Sn为{bn}的前n项和,当Sn最大时,求n的值.百度文库,精选习题试题习题,尽在百度答案精析1.解(1)因为2Sn=3n+3,所以2a1=3+3,故a1=3,当n>1时,2Sn-1=3n-1+3,此时2an=2Sn-2Sn-1=3n-3n-1=2×3n-1,即an=3n-1,显然a1不满足an=3n-1,所以an=3,n=1,3n-1,n>1.(2)因为anbn=log3an,所以b1=13,当n>1时,bn=31-nlog33n-1=(n-1)·31-n,所以T1=b1=13.当n>1时,Tn=b1+b2+b3+⋯+bn=13+[1×3-1+2×3-2+3×3-3+⋯+(n-1)×31-n],所以3Tn=1+[1×30+2×3-1+3×3-2+⋯+(n-1)×32-n],两式相减,得2Tn=23+(30+3-1+3-2+3-3+⋯+32-n)-(n-1)×31-n=23+1-31-n1-3-1-(n-1)×31-n=136-6n+32×3n,所以Tn=1312-6n+34×3n.经检验,n=1时也适合.综上可得Tn=1312-6n+34×3n.2.解(1)由题设知a1·a4=a2·a3=8.又a1+a4=9,可解得a1=1,a4=8或a1=8,a4=1(舍去).由a4=a1q3得公比q=2,故an=a1qn-1=2n-1(n∈N*).百度文库,精选习题试题习题,尽在百度(2)Sn=a11-qn1-q=2n-1,又bn=an+1SnSn+1=Sn+1-SnSnSn+1=1Sn-1Sn+1,所以Tn=b1+b2+⋯+bn=1S1-1S2+1S2-1S3+⋯+1Sn-1Sn+1=1S1-1Sn+1=1-12n+1-1.3.解(1)当n=1时,a1=S1=14a21+12a1-34.解得a1=3.又 4Sn=a2n+2an-3,①当n≥2时,4Sn-1=a2n-1+2an-1-3.②①-②,得4an=a2n-a2n-1+2(an-an-1),即a2n-a2n-1-2(an+an-1)=0.∴(an+an-1)(an-an-1-2)=0. an+an-1>0,∴an-an-1=2(n≥2),∴数列{an}是以3为首项,2为公差的等差数列.∴an=3+2(n-1)=2n+1.(2)Tn=3×21+5×22+⋯+(2n+1)·2n,③2Tn=3×22+5×23+⋯+(2n-1)·2n+(2n+1)2n+1,④④-③,得Tn=-3×21-2(22+23+⋯+2n)+(2n+1)2n+1=-6+8-2·2n+1+(2n+1)·2n+1=(2n-1)2n+1+2.4.解(1)由Sn=an+1-12,得Sn-1=an-12(n≥2),两式作差得an=an+1-an,即2an=an+1(n≥2),∴an+1an=2(n≥2),由a1=S1=a2-12=12,得a2=1,∴a2a1=2,∴数列{an}是首项为12,公比为2的等比数列.则an=12·2n-1=2n-2,Sn=an+1-12=2n-1-12.(2)bn=log2(2Sn+1)-2=log22n-2=n-2,∴cn·bn+3·bn+4=1+(n+1)(n+2)·2bn,百度文库,精选习题试题习题,尽在百度即cn(n+1)(n+2)=1+(n+1)(n+2)·2n-2,∴cn=1n+1n+2+2n-2=1n+1-1n+2+2n-2,∴Tn=(12-13)+(13-14)+⋯+(1n+1-1n+2)+(2-1+20+⋯+2n-2)=12-1n+2+121-2n1-2=12-1n+2-12+2n-1=2n-1-1n+2.由4Tn>2n+1-1504,得4(2n-1-1n+2)>2n+1-1504.即4n+2<1504,n>2014.∴使4Tn>2n+1-1504成...