（通用版）2020版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

专题突破练14求数列的通项及前n项和1.(2019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.(2019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=12nan,求数列bn的前n项和Sn.4.(2019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=Sn2ann,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(nN*).(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+1an,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anbn的前n项和Tn.7.设Sn是数列an的前n项和,an0,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:Tn0,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anbn=2n-12n.Tn=121+322+523+2n-12n,12Tn=122+323+524+2n-12n+1,-,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),当n=1时,4a1=a12+2a1,即a1=2.当n2时,4Sn-1=an-1(an-1+2).由-得4an=an2-an-12+2an-2an-1,即2(an+an-1)=(an+an-1)(an-an-1).an0,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+112.8.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