2022年高中数学 第二章 数列 5.docx

1、课时训练14数列求和一、分组求和1.若数列an的通项公式是an=(-1)n(3n-2),则a1+a2+a10=()A.15B.12C.-12D.-15答案:A解析:an=(-1)n(3n-2),则a1+a2+a10=-1+4-7+10-25+28=(-1+4)+(-7+10)+(-25+28)=35=15.2.已知数列an满足a1=1,an+1=an+n+2n(nN*),则an为()A.n(n-1)2+2n-1-1B.n(n-1)2+2n-1C.n(n+1)2+2n+1-1D.n(n-1)2+2n+1-1答案:B解析:an+1=an+n+2n,an+1-an=n+2n.an=a1+(a2-a1

2、)+(a3-a2)+(an-an-1)=1+(1+2)+(2+22)+(n-1)+2n-1=1+1+2+3+(n-1)+(2+22+2n-1)=1+(n-1)n2+2(1-2n-1)1-2=n(n-1)2+2n-1.3.(2015广东湛江高二期末,19)已知数列an为等差数列,a5=5,d=1;数列bn为等比数列,b4=16,q=2.(1)求数列an,bn的通项公式an,bn;(2)设cn=an+bn,求数列cn的前n项和Tn.解:(1)数列an为等差数列,a5=5,d=1,a1+4=5,解得a1=1,an=1+(n-1)1=n.数列bn为等比数列,b4=16,q=2,b123=16,解得b1

3、=2,bn=22n-1=2n.(2)cn=an+bn=n+2n,Tn=(1+2+3+n)+(2+22+23+2n)=n(n+1)2+2(1-2n)1-2=n2+n2+2n+1-2.二、裂项相消法求和4.数列an的通项公式an=11+2+3+n,则其前n项和Sn=()A.2nn+1B.n+12nC.(n+1)n2D.n2+n+2n+1答案:A解析:an=11+2+3+n=2n(n+1)=21n-1n+1,Sn=a1+a2+an=21-12+12-13+1n-1n+1=21-1n+1=2nn+1.5.113+135+157+1(2n-1)(2n+1)=.答案:n2n+1解析:1(2n-1)(2n+

4、1)=1212n-1-12n+1,113+135+157+1(2n-1)(2n+1)=121-13+13-15+15-17+12n-1-12n+1=121-12n+1=n2n+1.6.(2015山东省潍坊四县联考,17)等差数列an中,a1=3,其前n项和为Sn.等比数列bn的各项均为正数,b1=1,且b2+S2=12,a3=b3.(1)求数列an与bn的通项公式;(2)求数列1Sn的前n项和Tn.解:(1)设数列an的公差为d,数列bn的公比为q,由已知可得q+3+3+d=12,q2=3+2d,又q0,d=3,q=3,an=3+3(n-1)=3n,bn=3n-1.(2)由(1)知数列an中,

5、a1=3,an=3n,Sn=n(3+3n)2,1Sn=2n(3+3n)=231n-1n+1,Tn=231-12+12-13+1n-1n+1=231-1n+1=2n3(n+1).三、错位相减法求和7.数列22,422,623,2n2n,前n项的和为.答案:4-n+22n-1解析:设Sn=22+422+623+2n2n,12Sn=222+423+624+2n2n+1,-得1-12Sn=22+222+223+224+22n-2n2n+1=2-12n-1-2n2n+1.Sn=4-n+22n-1.8.(2015湖北高考,文19)设等差数列an的公差为d,前n项和为Sn,等比数列bn的公比为q,已知b1=

6、a1,b2=2,q=d,S10=100.(1)求数列an,bn的通项公式;(2)当d1时,记cn=anbn,求数列cn的前n项和Tn.解:(1)由题意有,10a1+45d=100,a1d=2,即2a1+9d=20,a1d=2,解得a1=1,d=2,或a1=9,d=29.故an=2n-1,bn=2n-1,或an=19(2n+79),bn=929n-1.(2)由d1,知an=2n-1,bn=2n-1,故cn=2n-12n-1,于是Tn=1+32+522+723+924+2n-12n-1,12Tn=12+322+523+724+925+2n-12n.-可得12Tn=2+12+122+12n-2-2n

7、-12n=3-2n+32n,故Tn=6-2n+32n-1.(建议用时:30分钟)1.数列an的通项公式是an=1n+n+1,若前n项和为10,则项数为()A.11B.99C.120D.121答案:C解析:an=1n+n+1=n+1-n,Sn=a1+a2+an=(2-1)+(3-2)+(n+1-n)=n+1-1,令n+1-1=10,得n=120.2.已知数列an的通项公式an=2n-12n,其前n项和Sn=32164,则项数n等于()A.13B.10C.9D.6答案:D解析:an=2n-12n=1-12n.Sn=n-121-12n1-12=n-1+12n=32164=5+164,n=6.3.数列

8、an的通项公式an=ncosn2,其前n项和为Sn,则S2 012等于()A.1 006B.2 012C.503D.0答案:A解析:函数y=cosn2的周期T=22=4,可分四组求和:a1+a5+a2009=0,a2+a6+a2010=-2-6-2010=503(-2-2010)2=-5031006,a3+a7+a2011=0,a4+a8+a2012=4+8+2012=503(4+2012)2=5031008.故S2012=0-5031006+0+5031008=503(-1006+1008)=1006.4.已知等比数列an的前n项和Sn=2n-1,则a12+a22+an2等于()A.(2n-

9、1)2B.13(2n-1)C.4n-1D.13(4n-1)答案:D解析:根据前n项和Sn=2n-1,可求出an=2n-1,由等比数列的性质可得an2仍为等比数列,且首项为a12,公比为q2,a12+a22+an2=1+22+24+22n-2=13(4n-1).5.已知数列an:12,13+23,14+24+34,15+25+35+45,那么数列bn=1anan+1前n项的和为()A.41-1n+1B.412-1n+1C.1-1n+1D.12-1n+1答案:A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+1

10、2-13+13-14+1n-1n+1=41-1n+1.6.如果lg x+lg x2+lg x10=110,那么lg x+lg2x+lg10x=.答案:2 046解析:由已知(1+2+10)lgx=110,55lgx=110.lgx=2.lgx+lg2x+lg10x=2+22+210=211-2=2046.7.已知等比数列an中,a1=3,a4=81.若数列bn满足bn=log3an,则数列1bnbn+1的前2 013项的和为.答案:20132014解析:a4a1=q3=27,q=3.an=a1qn-1=33n-1=3n.bn=log3an=n.1bnbn+1=1n(n+1)=1n-1n+1,数

11、列1bnbn+1的前2013项的和为:1-12+12-13+12013-12014=1-12014=20132014.8.已知等比数列an的各项都为正数,且当n3时,a4a2n-4=102n,则数列lg a1,2lg a2,22lg a3,23lg a4,2n-1lg an的前n项和Sn等于.答案:1+(n-1)2n解析:an是等比数列,a4a2n-4=an2=102n.an=10n,2n-1lgan=n2n-1.利用错位相减法求得Sn=1+(n-1)2n.9.正项数列an满足:an2-(2n-1)an-2n=0.(1)求数列an的通项公式an;(2)令bn=1(n+1)an,求数列bn的前n

12、项和Tn.解:(1)由an2-(2n-1)an-2n=0,得(an-2n)(an+1)=0.由于an是正项数列,所以an=2n.(2)由an=2n,bn=1(n+1)an,则bn=12n(n+1)=121n-1n+1,Tn=121-12+12-13+1n-1-1n+1n-1n+1=121-1n+1=n2(n+1).10.已知数列an的前n项和为Sn,且Sn=2n2+n,nN*,数列bn满足an=4log2bn+3,nN*.(1)求an,bn;(2)求数列anbn的前n项和Tn.解:(1)由Sn=2n2+n,得当n=1时,a1=S1=3;当n2时,an=Sn-Sn-1=4n-1.当n=1时,41-1=3.所以an=4n-1,nN*.由4n-1=an=4log2bn+3,得bn=2n-1,nN*.(2)由(1)知anbn=(4n-1)2n-1,nN*.所以Tn=3+72+1122+(4n-1)2n-1,2Tn=32+722+(4n-5)2n-1+(4n-1)2n,所以2Tn-Tn=(4n-1)2n-3+4(2+22+2n-1)=(4n-5)2n+5.故Tn=(4n-5)2n+5,nN*.