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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
2、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(n∈N*).
(1)證明:數(shù)列{a
3、n+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.
4、
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(n∈N*),-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
5、,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=
6、(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
7、+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)·
8、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(n∈N*),
∴an+2-an+1=2(an+1-an)(n∈N*),
∴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(n∈N*),
∴an=(an-an-1)+(an-1-an-2)+…+(a2-a1)+a1=2n-1+2n-2+…+2+1=2n-1(n∈N*).
9、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,
∴l(xiāng)og2bn=-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,
10、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
當(dāng)n=1時(shí),4a1=a12+2a1,即a1=2.
當(dāng)n≥2時(shí),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)證明∵bn=1(an-1)(an+1)=1(2n-1)(2n+
11、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).
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（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
