2023学年高考数学大二轮复习专题突破练14求数列的通项及前n项和理2.docx

专题突破练14　求数列的通项及前n项和 1.(2023江西宜春高三上学期期末)已知等差数列{an}的前n项和为Sn,且a2+a6=10,S5=20. (1)求an与Sn; (2)设数列{cn}满足cn=1Sn-n,求{cn}的前n项和Tn. 2.(2023吉林高中高三上学期期末考试)在递增的等比数列{an}中,a2=6,且4(a3-a2)=a4-6. (1)求{an}的通项公式; (2)若bn=an+2n-1,求数列{bn}的前n项和Sn. 3.已知数列{an}满足a1=12,an+1=an2an+1. (1)证明数列1an是等差数列,并求{an}的通项公式; (2)若数列{bn}满足bn=12n·an,求数列{bn}的前n项和Sn. 4.(2023辽宁朝阳重点高中高三第四次模拟)已知等差数列{an}的前n项和为Sn,满足S3=12,且a1,a2,a4成等比数列. (1)求an及Sn; (2)设bn=Sn·2ann,数列{bn}的前n项和为Tn,求Tn. 5.已知数列{an}满足a1=1,a2=3,an+2=3an+1-2an(n∈N*). (1)证明:数列{an+1-an}是等比数列; (2)求数列{an}的通项公式和前n项和Sn. 6.已知等差数列{an}满足:an+1>an,a1=1,该数列的前三项分别加上1,1,3后成等比数列,an+2log2bn=-1. (1)求数列{an},{bn}的通项公式; (2)求数列{an·bn}的前n项和Tn. 7.设Sn是数列{an}的前n项和,an>0,且4Sn=an(an+2). (1)求数列{an}的通项公式; (2)设bn=1(an-1)(an+1),Tn=b1+b2+…+bn,求证:Tn<12. 8.(2023山东淄博部分学校高三阶段性诊断考试)已知等比数列{an}的前n项和为Sn(n∈N*),-2S2,S3,4S4成等差数列,且a2+2a3+a4=116. (1)求数列{an}的通项公式; (2)若bn=-(n+2)log2|an|,求数列1bn的前n项和Tn. 　　　　　　　　　　　　　　　　 2023学年参考答案 专题突破练14　求数列的 通项及前n项和 1.解(1)设等差数列公差为d, S5=5(a1+a5)2=5a3=20,故a3=4, a2+a6=2a4=10,故a4=5, ∴d=1,an=a3+d(n-3)=n+1, 易得a1=2, ∴Sn=n2(a1+an)=n2(2+n+1)=n(n+3)2. (2)由(1)知Sn=n(n+3)2, 则cn=1Sn-n=2n2+n=21n-1n+1, 则Tn=21-12+12-13+13-14+…+1n-1n+1=21-1n+1=2nn+1. 2.解(1)设公比为q,由4(a3-a2)=a4-6,得4(6q-6)=6q2-6, 化简得q2-4q+3=0,解得q=3或q=1, 因为等比数列{an}是递增的,所以q=3,a1=2, 所以an=2×3n-1. (2)由(1)得bn=2×3n-1+2n-1, 所以Sn=(2+6+18+…+2×3n-1)+(1+3+5+…+2n-1), 则Sn=2×(1-3n)1-3+n(1+2n-1)2, 所以Sn=3n-1+n2. 3.(1)证明∵an+1=an2an+1, ∴1an+1-1an=2, ∴1an是等差数列, ∴1an=1a1+(n-1)×2=2+2n-2=2n,即an=12n. (2)解∵bn=12n·an=2n2n, ∴Sn=b1+b2+…+bn=1+22+322+…+n2n-1, 则12Sn=12+222+323+…+n2n, 两式相减得12Sn=1+12+122+123+…+12n-1-n2n=21-12n-n2n,∴Sn=4-2+n2n-1. 4.解(1)设等差数列{an}的公差为d, 因为S3=12,且a1,a2,a4成等比数列, 所以有S3=3a2=12,a22=a1a4, 即a1+d=4,(a1+d)2=a1(a1+3d), 解得a1=2,d=2. 所以an=a1+(n-1)d=2n,Sn=n(a1+an)2=n2+n. (2)由(1)可得 bn=Sn·2ann=n(n+1)·22nn =(n+1)·4n, 因为数列{bn}的前n项和为Tn, 所以Tn=b1+b2+b3+…+bn=2×4+3×42+4×43+…+(n+1)·4n,因此,4Tn=2×42+3×43+4×44+…+(n+1)·4n+1, 两式作差,得-3Tn=2×4+42+43+44+…+4n-(n+1)·4n+1, 整理得Tn=(3n+2)·4n+1-89. 5.(1)证明∵an+2=3an+1-2an(n∈N*), ∴an+2-an+1=2(an+1-an)(n∈N*), ∴an+2-an+1an+1-an=2. ∵a1=1,a2=3,∴数列{an+1-an}是以a2-a1=2为首项,公比为2的等比数列. (2)解由(1)得,an+1-an=2n(n∈N*), ∴an=(an-an-1)+(an-1-an-2)+…+(a2-a1)+a1=2n-1+2n-2+…+2+1=2n-1(n∈N*). Sn=(2-1)+(22-1)+(23-1)+…+(2n-1)=(2+22+23+…+2n)-n=2(1-2n)1-2-n=2n+1-2-n. 6.解(1)设等差数列{an}的公差为d,且d>0,由a1=1,a2=1+d,a3=1+2d,分别加上1,1,3后成等比数列,得(2+d)2=2(4+2d),解得d=2,∴an=1+(n-1)×2=2n-1. ∵an+2log2bn=-1, ∴log2bn=-n,即bn=12n. (2)由(1)得an·bn=2n-12n.Tn=121+322+523+…+2n-12n,① 12Tn=122+323+524+…+2n-12n+1,② ①-②,得12Tn=12+2122+123+124+…+12n-2n-12n+1. ∴Tn=1+1-12n-11-12-2n-12n=3-12n-2-2n-12n=3-2n+32n. 7.(1)解4Sn=an(an+2),① 当n=1时,4a1=a12+2a1,即a1=2. 当n≥2时,4Sn-1=an-1(an-1+2).② 由①-②得4an=an2-an-12+2an-2an-1,即2(an+an-1)=(an+an-1)·(an-an-1). ∵an>0,∴an-an-1=2, ∴an=2+2(n-1)=2n. (2)证明∵bn=1(an-1)(an+1)=1(2n-1)(2n+1) =1212n-1-12n+1, ∴Tn=b1+b2+…+bn=121-13+13-15+…+12n-1-12n+1=121-12n+1<12. 8.解(1)设等比数列{an}的公比为q. 由-2S2,S3,4S4成等差数列知, 2S3=-2S2+4S4, 所以2a4=-a3,即q=-12. 又a2+2a3+a4=116, 所以a1q+2a1q2+a1q3=116, 所以a1=-12. 所以等差数列{an}的通项公式an=-12n. (2)由(1)知 bn=-(n+2)log2-12n =n(n+2), 所以1bn=1n(n+2)=121n-1n+2. 所以数列1bn的前n项和: Tn=121-13+12-14+13-15+…+1n-1-1n+1+1n-1n+2 =121+12-1n+1-1n+2 =34-2n+32(n+1)(n+2). 所以数列1bn的前n项和Tn=34-2n+32(n+1)(n+2). 12