（通用版）2020版高考數(shù)學(xué)大二輪復(fù)習(xí) 專題突破練14 求數(shù)列的通項(xiàng)及前n項(xiàng)和 理

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1、專題突破練14求數(shù)列的通項(xiàng)及前n項(xiàng)和1.(2019江西宜春高三上學(xué)期期末)已知等差數(shù)列an的前n項(xiàng)和為Sn,且a2+a6=10,S5=20.(1)求an與Sn;(2)設(shè)數(shù)列cn滿足cn=1Sn-n,求cn的前n項(xiàng)和Tn.2.(2019吉林高中高三上學(xué)期期末考試)在遞增的等比數(shù)列an中,a2=6,且4(a3-a2)=a4-6.(1)求an的通項(xiàng)公式;(2)若bn=an+2n-1,求數(shù)列bn的前n項(xiàng)和Sn.3.已知數(shù)列an滿足a1=12,an+1=an2an+1.(1)證明數(shù)列1an是等差數(shù)列,并求an的通項(xiàng)公式;(2)若數(shù)列bn滿足bn=12nan,求數(shù)列bn的前n項(xiàng)和Sn.4.(2019遼寧朝

2、陽重點(diǎn)高中高三第四次模擬)已知等差數(shù)列an的前n項(xiàng)和為Sn,滿足S3=12,且a1,a2,a4成等比數(shù)列.(1)求an及Sn;(2)設(shè)bn=Sn2ann,數(shù)列bn的前n項(xiàng)和為Tn,求Tn.5.已知數(shù)列an滿足a1=1,a2=3,an+2=3an+1-2an(nN*).(1)證明:數(shù)列an+1-an是等比數(shù)列;(2)求數(shù)列an的通項(xiàng)公式和前n項(xiàng)和Sn.6.已知等差數(shù)列an滿足:an+1an,a1=1,該數(shù)列的前三項(xiàng)分別加上1,1,3后成等比數(shù)列,an+2log2bn=-1.(1)求數(shù)列an,bn的通項(xiàng)公式;(2)求數(shù)列anbn的前n項(xiàng)和Tn.7.設(shè)Sn是數(shù)列an的前n項(xiàng)和,an0,且4Sn=an

3、(an+2).(1)求數(shù)列an的通項(xiàng)公式;(2)設(shè)bn=1(an-1)(an+1),Tn=b1+b2+bn,求證:Tn0,由a1=1,a2=1+d,a3=1+2d,分別加上1,1,3后成等比數(shù)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-

4、12n-2-2n-12n=3-2n+32n.7.(1)解4Sn=an(an+2),當(dāng)n=1時(shí),4a1=a12+2a1,即a1=2.當(dāng)n2時(shí),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)證明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)設(shè)等比數(shù)列an的公比為q.由-2S2,S3,4S4成等差數(shù)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.所以等差數(shù)列an的通項(xiàng)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.所以數(shù)列1bn的前n項(xiàng)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).所以數(shù)列1bn的前n項(xiàng)和Tn=34-2n+32(n+1)(n+2).12