（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

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=12n·an,求數(shù)列bn的前n項(xiàng)和Sn.4.(2019遼寧朝陽重點(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=Sn·2ann,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+1>an,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an·bn的前n項(xiàng)和Tn.7.設(shè)Sn是數(shù)列an的前n項(xiàng)和,an>0,且4Sn=an(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:Tn<12.8.(2019山東淄博部分學(xué)校高三階段性診斷考試)已知等比數(shù)列an的前n項(xiàng)和為Sn(nN*),-2S2,S3,4S4成等差數(shù)列,且a2+2a3+a4=116.(1)求數(shù)列an的通項(xiàng)公式;(2)若bn=-(n+2)log2|an|,求數(shù)列1bn的前n項(xiàng)和Tn.參考答案專題突破練14求數(shù)列的通項(xiàng)及前n項(xiàng)和1.解(1)設(shè)等差數(shù)列公差為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,則cn=1Sn-n=2n2+n=21n-1n+1,則Tn=21-12+12-13+13-14+1n-1n+1=21-1n+1=2nn+1.2.解(1)設(shè)公比為q,由4(a3-a2)=a4-6,得4(6q-6)=6q2-6,化簡(jiǎn)得q2-4q+3=0,解得q=3或q=1,因?yàn)榈缺葦?shù)列an是遞增的,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),則Sn=2×(1-3n)1-3+n(1+2n-1)2,所以Sn=3n-1+n2.3.(1)證明an+1=an2an+1,1an+1-1an=2,1an是等差數(shù)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,則12Sn=12+222+323+n2n,兩式相減得12Sn=1+12+122+123+12n-1-n2n=21-12n-n2n,Sn=4-2+n2n-1.4.解(1)設(shè)等差數(shù)列an的公差為d,因?yàn)镾3=12,且a1,a2,a4成等比數(shù)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?yàn)閿?shù)列bn的前n項(xiàng)和為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+1,兩式作差,得-3Tn=2×4+42+43+44+4n-(n+1)·4n+1,整理得Tn=(3n+2)·4n+1-89.5.(1)證明an+2=3an+1-2an(nN*),an+2-an+1=2(an+1-an)(nN*),an+2-an+1an+1-an=2.a1=1,a2=3,數(shù)列an+1-an是以a2-a1=2為首項(xiàng),公比為2的等比數(shù)列.(2)解由(1)得,an+1-an=2n(nN*),an=(an-an-1)+(an-1-an-2)+(a2-a1)+a1=2n-1+2n-2+2+1=2n-1(nN*).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)設(shè)等差數(shù)列an的公差為d,且d>0,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an·bn=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),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).an>0,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+1<12.8.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