2019版数学人教A版必修5训练：第二章　习题课（二）　数列求和 WORD版含解析.docx

1、习题课(二)数列求和课时过关能力提升基础巩固1设数列an的前n项和为Sn,如果an=1(2n-1)(2n+1),那么S5等于().A.12B.511C.49D.59解析:an=1(2n-1)(2n+1)=1212n-1-12n+1,S5=121-13+13-15+15-17+17-19+19-111=511.答案:B2若数列an的通项公式为an=2n+2n-1,则数列an的前n项和为().A.2n+n2-1B.2n+1+n2-1C.2n+1+n2-2D.2n+n-2解析:Sn=(2+22+2n)+(1+3+5+2n-1)=2(1-2n)1-2+(1+2n-1)n2=2n+1-2+n2.答案:C

2、3数列an的通项公式an=1n+n+1,若该数列前n项的和为10,则项数为().A.11B.99C.120D.121解析:an=1n+n+1=n+1-n,Sn=n+1-1=10,n=120.答案:C4已知数列an中,a1=1,前n项和为Sn,且点P(an,an+1)(nN*)在直线x-y+1=0上,则1S1+1S2+1S3+1Sn等于()A.2nn+1B.2n(n+1)C.n(n+1)2D.n2(n+1)解析:由题意,an-an+1+1=0,an+1-an=1.an为等差数列,且a1=1,d=1,an=1+(n-1)1=n,Sn=n(n+1)2.1Sn=2n(n+1)=21n-1n+1.1S1

3、+1S2+1Sn=21-12+12-13+1n-1n+1=2nn+1.答案:A5数列12,24,38,n2n,的前n项和为 .解析:Sn=12+222+323+n2n,12Sn=122+223+n-12n+n2n+1,-,得12Sn=12+122+123+12n-n2n+1=121-12n1-12-n2n+1=1-12n-n2n+1,Sn=2-12n-1-n2n.答案:Sn=2-12n-1-n2n6已知an=ln1+1n(nN*),则数列an的前n项和为Sn=.解析:an=lnn+1n=ln(n+1)-ln n,Sn=(ln 2-ln 1)+(ln 3-ln 2)+(ln 4-ln 3)+ln

4、(n+1)-ln n=ln(n+1)-ln 1=ln(n+1).答案:ln(n+1)7设数列an的前n项和为Sn.已知S2=4,an+1=2Sn+1,nN*.(1)求通项公式an;(2)求数列|an-n-2|的前n项和.解(1)由题意得a1+a2=4,a2=2a1+1,则a1=1,a2=3.又当n2时,由an+1-an=(2Sn+1)-(2Sn-1+1)=2an,得an+1=3an.所以,数列an的通项公式为an=3n-1,nN*.(2)设bn=|3n-1-n-2|,nN*,b1=2,b2=1.当n3时,由于3n-1n+2,故bn=3n-1-n-2,n3.设数列bn的前n项和为Tn,则T1=2

5、,T2=3.当n3时,Tn=3+9(1-3n-2)1-3-(n+7)(n-2)2=3n-n2-5n+112,所以Tn=2,n=1,3n-n2-5n+112,n2,nN*.8已知等差数列an的前n项和Sn满足S3=0,S5=-5.(1)求数列an的通项公式;(2)求数列1a2n-1a2n+1的前n项和.解(1)设等差数列an的公差为d,则Sn=na1+n(n-1)2d.由已知可得3a1+3d=0,5a1+10d=-5,解得a1=1,d=-1.故数列an的通项公式为an=2-n.(2)由(1)知1a2n-1a2n+1=1(3-2n)(1-2n)=1212n-3-12n-1,从而数列1a2n-1a2

6、n+1的前n项和为121-1-11+11-13+12n-3-12n-1=n1-2n.9已知an是等差数列,bn是等比数列,且b2=3,b3=9,a1=b1,a14=b4.(1)求an的通项公式;(2)设cn=an+bn,求数列cn的前n项和.解(1)等比数列bn的公比q=b3b2=93=3,所以b1=b2q=1,b4=b3q=27.设等差数列an的公差为d.因为a1=b1=1,a14=b4=27,所以1+13d=27,即d=2.所以an=2n-1(n=1,2,3,).(2)由(1)知,an=2n-1,bn=3n-1.因此cn=an+bn=2n-1+3n-1.从而数列cn的前n项和Sn=1+3+

7、(2n-1)+1+3+3n-1=n(1+2n-1)2+1-3n1-3=n2+3n-12.能力提升1数列an,bn都是等差数列,a1=5,b1=7,且a20+b20=60,则an+bn的前20项和为().A.700B.710C.720D.730解析:数列an+bn也是等差数列,其首项为12,第20项为60,所以其前20项和为20(a1+b1+a20+b20)2=20(12+60)2=720.答案:C2已知数列an的通项公式an=2n-12n,其前n项和Sn=32164,则n的值为().A.13B.10C.9D.6解析:an=2n-12n=1-12n,Sn=n-121-12n1-12=n-1+12

8、n=32164=5+164,n=6.答案:D3已知数列an:12,13+23,14+24+34,15+25+35+45,则数列1anan+1的前n项和为().A.41-1n+1B.412-1n+1C.1-1n+1D.12-1n+1解析:an=1+2+3+nn+1=n(n+1)2n+1=n2,bn=1anan+1=4n(n+1)=41n-1n+1.Sn=41-12+12-13+13-14+1n-1n+1=41-1n+1.答案:A4已知函数f(n)=n2,n为奇数,-n2,n为偶数,且an=f(n)+f(n+1),则a1+a2+a3+a100等于()A.0B.100C.-100D.10 200解析

9、:由题意,得a1+a2+a3+a100=12-22-22+32+32-42-42+52+992-1002-1002+1012=-(1+2)+(3+2)-(4+3)+-(99+100)+(101+100)=-(1+2+99+100)+(2+3+100+101)=-50101+50103=100.故选B.答案:B5在数列an中,an=nsinn2+cosn2,前n项和为Sn,则S100=.解析:易知a1=1,a2=-2,a3=-3,a4=4,a1+a2+a3+a4=0.又sinn2+cosn2的周期为4,an+an+1+an+2+an+3=0,S100=0.答案:06在有限数列an中,Sn为an的

10、前n项和,把S1+S2+Snn称为数列an的“优化和”.若数列a1,a2,a3,a2 019的“优化和”为2 020,则数列1,a1,a2,a3,a2 019的“优化和”为.解析:设数列1,a1,a2,a3,a2 019的前n项和为Tn,则T1=1,T2=S1+1,T3=S2+1,T4=S3+1,T2 019=S2 018+1,T2 020=S2 019+1,于是T1+T2+T3+T2 020=2 020+S1+S2+S2 019.S1+S2+S2 0192 019=2 020,S1+S2+S2 019=2 0192 020.T1+T2+T3+T2 020=2 020+2 0192 020=2

11、 0202,其优化和为2 02022 020=2 020.答案:2 0207等比数列an的前n项和为Sn,已知对任意的nN*,点(n,Sn)均在函数y=bx+r(b0,且b1,b,r均为常数)的图象上.(1)求r的值;(2)当b=2时,记bn=n+14an(nN*),求数列bn的前n项和Tn.解(1)由题意,得Sn=bn+r,当n2时,Sn-1=bn-1+r,an=Sn-Sn-1=bn-1(b-1).b0,且b1,当n2时,数列an是以b为公比的等比数列.又a1=b+r,a2=b(b-1),a2a1=b,即b(b-1)b+r=b,解得r=-1.(2)由(1)知,an=(b-1)bn-1=2n-1,nN*,bn=n+142n-1=n+12n+1.Tn=222+323+424+n+12n+1,12Tn=223+324+n2n+1+n+12n+2,两式相减,得12Tn=222+123+124+12n+1-n+12n+2=12+1231-12n-11-12-n+12n+2=34-12n+1-n+12n+2,故Tn=32-12n-n+12n+1=32-n+32n+1.