1、考点规范练30数列求和考点规范练B册第20页基础巩固组1.数列112,314,518,7116,(2n-1)+12n,的前n项和Sn的值等于()A.n2+1-12nB.2n2-n+1-12nC.n2+1-12n-1D.n2-n+1-12n答案:A解析:该数列的通项公式为an=(2n-1)+12n,则Sn=1+3+5+(2n-1)+12+122+12n=n2+1-12n.2.(2015云南曲靖一模)122-1+132-1+142-1+1(n+1)2-1的值为()A.n+12(n+2)B.34-n+12(n+2)C.34-121n+1+1n+2D.32-1n+1+1n+2导学号32470776答案
2、:C解析:1(n+1)2-1=1n2+2n=1n(n+2)=121n-1n+2,122-1+132-1+142-1+1(n+1)2-1=121-13+12-14+13-15+1n-1n+2=1232-1n+1-1n+2=34-121n+1+1n+2.3.已知数列an:12,13+23,14+24+34,110+210+310+910,若bn=1anan+1,那么数列bn的前n项和Sn等于()A.nn+1B.4nn+1C.3nn+1D.5nn+1导学号32470777答案:B解析:易得an=1+2+3+nn+1=n2,bn=1anan+1=4n(n+1)=41n-1n+1.Sn=41-12+12
3、-13+1n-1n+1=41-1n+1=4nn+1.4.已知函数f(x)=xa的图象过点(4,2),令an=1f(n+1)+f(n),nN*.记数列an的前n项和为Sn,则S2 016等于()A.2016-1B.2016+1C.2017-1D.2017+1答案:C解析:由f(4)=2,可得4a=2,解得a=12,则f(x)=x12.an=1f(n+1)+f(n)=1n+1+n=n+1-n,S2 016=a1+a2+a3+a2 016=(2-1)+(3-2)+(4-3)+(2017-2016)=2017-1.5.数列an满足an+1+(-1)nan=2n-1,则an的前60项和为()A.3 69
4、0B.3 660C.1 845D.1 830答案:D解析:an+1+(-1)nan=2n-1,当n=2k(kN*)时,a2k+1+a2k=4k-1,当n=2k+1(kN)时,a2k+2-a2k+1=4k+1,+得:a2k+a2k+2=8k.则a2+a4+a6+a8+a60=(a2+a4)+(a6+a8)+(a58+a60)=8(1+3+29)=815(1+29)2=1 800.由得a2k+1=a2k+2-(4k+1),a1+a3+a5+a59=a2+a4+a60-4(0+1+2+29)+30=1 800-430292+30=30,a1+a2+a60=1 800+30=1 830.6.32-1+
5、42-2+52-3+(n+2)2-n=.答案:4-n+42n解析:设Sn=32-1+42-2+52-3+(n+2)2-n,则12Sn=32-2+42-3+(n+1)2-n+(n+2)2-n-1.-,得12Sn=32-1+2-2+2-3+2-n-(n+2)2-n-1=22-1+2-1+2-2+2-3+2-n-(n+2)2-n-1=1+2-1(1-2-n)1-2-1-(n+2)2-n-1=2-(n+4)2-n-1.故Sn=4-n+42n.7.已知在等比数列an中,a1=3,a4=81,若数列bn满足bn=log3an,则数列1bnbn+1的前n项和Sn=.导学号32470778答案:nn+1解析:
6、设等比数列an的公比为q,则a4a1=q3=27,解得q=3.an=a1qn-1=33n-1=3n,故bn=log3an=n,1bnbn+1=1n(n+1)=1n-1n+1,则数列1bnbn+1的前n项和为1-12+12-13+1n-1n+1=1-1n+1=nn+1.8.(2015长春模拟)已知等差数列an满足:a5=9,a2+a6=14.(1)求an的通项公式;(2)若bn=an+qan(q0),求数列bn的前n项和Sn.解:(1)在等差数列an中,a5=9,a2+a6=2a4=14,a4=7,其公差d=a5-a4=2,an=a4+(n-4)d=7+2(n-4)=2n-1.(2)bn=an+
7、qan(q0),Sn=b1+b2+bn=(a1+a2+an)+(qa1+qa2+qan)=1+3+5+(2n-1)+(q1+q3+q2n-1)=n2+(q1+q3+q2n-1).若q=1,Sn=n2+n;若q1,Sn=n2+q(1-q2n)1-q2.9.(2015湖北,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=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
8、,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-12n=3-2n+32n,故Tn=6-2n+32n-1.导学号3247077910.(2015山东,文19)已知数列an是首项为正数的等差数列,数列1anan+1的前n项和为n2n+1.(1)求数列an的通项公式;(2)设bn=(an+1)2a
9、n,求数列bn的前n项和Tn.解:(1)设数列an的公差为d.令n=1,得1a1a2=13,所以a1a2=3.令n=2,得1a1a2+1a2a3=25,所以a2a3=15.解得a1=1,d=2,所以an=2n-1.(2)由(1)知bn=(an+1)2an=2n22n-1=n4n,所以Tn=141+242+n4n,所以4Tn=142+243+n4n+1,两式相减,得-3Tn=41+42+4n-n4n+1=4(1-4n)1-4-n4n+1=1-3n34n+1-43.所以Tn=3n-194n+1+49=4+(3n-1)4n+19.11.在数列an中,a1=1,当n2时,其前n项和Sn满足Sn2=an
10、Sn-12.(1)求Sn的表达式;(2)设bn=Sn2n+1,求数列bn的前n项和Tn.解:(1)Sn2=anSn-12,又an=Sn-Sn-1(n2),Sn2=(Sn-Sn-1)Sn-12,即2Sn-1Sn=Sn-1-Sn.由题意得Sn-1Sn0,式两边同除以Sn-1Sn,得1Sn-1Sn-1=2,数列1Sn是首项为1S1=1a1=1,公差为2的等差数列.1Sn=1+2(n-1)=2n-1,Sn=12n-1.(2)bn=Sn2n+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=n2n+1
11、.能力提升组12.已知正项数列an,bn满足a1=3,a2=6,bn是等差数列,且对任意正整数n,都有bn,an,bn+1成等比数列.(1)求数列bn的通项公式;(2)设Sn=1a1+1a2+1an,试比较2Sn与2-bn+12an+1的大小.解:(1)对任意正整数n,都有bn,an,bn+1成等比数列,且an,bn都为正项数列,an=bnbn+1.a1=b1b2=3,a2=b2b3=6.又bn是等差数列,b1+b3=2b2,解得b1=2,b2=322.bn=22(n+1).(2)由(1)可得an=bnbn+1=(n+1)(n+2)2,则1an=2(n+1)(n+2)=21n+1-1n+2,S
12、n=212-13+13-14+1n+1-1n+2=1-2n+2.2Sn=2-4n+2.又2-bn+12an+1=2-n+2n+3,2Sn-2-bn+12an+1=n+2n+3-4n+2=n2-8(n+2)(n+3).当n=1,2时,2Sn2-bn+12an+1.导学号3247078013.已知数列an是等差数列,bn是等比数列,其中a1=b1=1,a2b2,且b2为a1,a2的等差中项,a2为b2,b3的等差中项.(1)求数列an与bn的通项公式;(2)记cn=1n(a1+a2+an)(b1+b2+bn),求数列cn的前n项和Sn.解:(1)设an的公差为d,bn的公比为q.由2b2=a1+a2,2a2=b2+b3,得q=1,d=0或q=2,d=2.又a2b2,q=2,d=2,an=2n-1,bn=2n-1.(2)设an的前n项和为Mn,bn的前n项和为n,由(1)得Mn=a1+a2+an=a1+an2n=n2,n=b1+b2+bn=b1(1-qn)1-q=1-2n1-2=2n-1,故cn=1nn2(2n-1)=n2n-n,Sn=(121+222+n2n)-(1+2+n).令Tn=121+222+323+n2n,2Tn=122+223+324+n2n+1,由-,得Tn=(n-1)2n+1+2.Sn=(n-1)2n+1-n(n+1)2+2.导学号324707815