1、考点规范练32数列求和基础巩固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.数列an满足a1=1,且对任意的nN*都有an+1=a1+an+n,则1an的前100项和为()A.100101B.99100C.101100D.200101答案:D解析:an+1=a1+an+n,a1=1,an+1-an=1+n.an-an-1=
2、n(n2).an=(an-an-1)+(an-1-an-2)+(a2-a1)+a1=n+(n-1)+2+1=n(n+1)2.1an=2n(n+1)=21n-1n+1.1an的前100项和为21-12+12-13+1100-1101=21-1101=200101.故选D.3.已知数列an满足an+1-an=2,a1=-5,则|a1|+|a2|+|a6|=()A.9B.15C.18D.30答案:C解析:an+1-an=2,a1=-5,数列an是首项为-5,公差为2的等差数列.an=-5+2(n-1)=2n-7.数列an的前n项和Sn=n(-5+2n-7)2=n2-6n.令an=2n-70,解得n7
3、2.当n3时,|an|=-an;当n4时,|an|=an.|a1|+|a2|+|a6|=-a1-a2-a3+a4+a5+a6=S6-2S3=62-66-2(32-63)=18.4.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,S2016=a1+a2+a3+a2016=(2-1)+(3-2)+(4-3)+(20
4、17-2016)=2017-1.5.已知数列an满足an+1+(-1)nan=2n-1,则an的前60项和为()A.3 690B.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=1800.由得a2k+1=a2k+2-(4k+1),a1+a3+a5+a59=a2+a4+a60-4(0+1
5、+2+29)+30=1800-430292+30=30,a1+a2+a60=1800+30=1830.6.已知在数列an中,a1=1,且an+1=an2an+1,若bn=anan+1,则数列bn的前n项和Sn为()A.2n2n+1B.n2n+1C.2n2n-1D.2n-12n+1答案:B解析:由an+1=an2an+1,得1an+1=1an+2,数列1an是以1为首项,2为公差的等差数列,1an=2n-1,又bn=anan+1,bn=1(2n-1)(2n+1)=1212n-1-12n+1,Sn=1211-13+13-15+12n-1-12n+1=n2n+1,故选B.7.已知等差数列an,a5=
6、2.若函数f(x)=sin 2x+1,记yn=f(an),则数列yn的前9项和为.答案:9解析:由题意,得yn=sin(2an)+1,所以数列yn的前9项和为sin2a1+sin2a2+sin2a3+sin2a8+sin2a9+9.由a5=2,得sin2a5=0.a1+a9=2a5=,2a1+2a9=4a5=2,2a1=2-2a9,sin2a1=sin2-2a9=-sin2a9.由倒序相加可得12(sin2a1+sin2a2+sin2a3+sin2a8+sin2a9+sin2a1+sin2a2+sin2a3+sin2a8+sin2a9)=0,y1+y2+y3+y8+y9=9.8.在数列an中,
7、a1=3,an的前n项和Sn满足Sn+1=an+n2.(1)求数列an的通项公式;(2)设数列bn满足bn=(-1)n+2an,求数列bn的前n项和Tn.解:(1)由Sn+1=an+n2,得Sn+1+1=an+1+(n+1)2,-,得an=2n+1.a1=3满足上式,所以数列an的通项公式为an=2n+1.(2)由(1)得bn=(-1)n+22n+1,所以Tn=b1+b2+bn=(-1)+(-1)2+(-1)n+(23+25+22n+1)=(-1)1-(-1)n1-(-1)+23(1-4n)1-4=(-1)n-12+83(4n-1).9.设等差数列an的公差为d,前n项和为Sn,等比数列bn的
8、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,得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+
9、12n-2-2n-12n=3-2n+32n,故Tn=6-2n+32n-1.10.已知Sn为数列an的前n项和,an0,an2+2an=4Sn+3.(1)求an的通项公式;(2)设bn=1anan+1,求数列bn的前n项和.解:(1)由an2+2an=4Sn+3,可知an+12+2an+1=4Sn+1+3.两式相减可得an+12-an2+2(an+1-an)=4an+1,即2(an+1+an)=an+12-an2=(an+1+an)(an+1-an).由于an0,可得an+1-an=2.又a12+2a1=4a1+3,解得a1=-1(舍去),a1=3.所以an是首项为3,公差为2的等差数列,故an
10、的通项公式为an=2n+1.(2)由an=2n+1可知bn=1anan+1=1(2n+1)(2n+3)=1212n+1-12n+3.设数列bn的前n项和为Tn,则Tn=b1+b2+bn=1213-15+15-17+12n+1-12n+3=n3(2n+3).11.已知各项均为正数的数列an的前n项和为Sn,满足an+12=2Sn+n+4,a2-1,a3,a7恰为等比数列bn的前3项.(1)求数列an,bn的通项公式;(2)若cn=(-1)nlog2bn-1anan+1,求数列cn的前n项和Tn.解:(1)因为an+12=2Sn+n+4,所以an2=2Sn-1+n-1+4(n2).两式相减,得an
11、+12-an2=2an+1,所以an+12=an2+2an+1=(an+1)2.因为an是各项均为正数的数列,所以an+1=an+1,即an+1-an=1.又a32=(a2-1)a7,所以(a2+1)2=(a2-1)(a2+5),解得a2=3,a1=2,所以an是以2为首项,1为公差的等差数列,所以an=n+1.由题意知b1=2,b2=4,b3=8,故bn=2n.(2)由(1)得cn=(-1)nlog22n-1(n+1)(n+2)=(-1)nn-1(n+1)(n+2),故Tn=c1+c2+cn=-1+2-3+(-1)nn-123+134+1(n+1)(n+2).设Fn=-1+2-3+(-1)n
12、n.则当n为偶数时,Fn=(-1+2)+(-3+4)+-(n-1)+n=n2;当n为奇数时,Fn=Fn-1+(-n)=n-12-n=-(n+1)2.设Gn=123+134+1(n+1)(n+2),则Gn=12-13+13-14+1n+1-1n+2=12-1n+2.所以Tn=n-12+1n+2,n为偶数,-n+22+1n+2,n为奇数.能力提升12.今要在一个圆周上标出一些数,第一次先把圆周二等分,在这两个分点处分别标上1,如图所示;第二次把两段半圆弧二等分,在这两个分点处分别标上2,如图所示;第三次把4段圆弧二等分,并在这4个分点处分别标上3,如图所示.如此继续下去,当第n次标完数以后,这个圆
13、周上所有已标出的数的总和是.答案:(n-1)2n+2解析:由题意可得,第n次标完后,圆周上所有已标出的数的总和为Tn=1+1+22+322+n2n-1.设S=1+22+322+n2n-1,则2S=2+222+(n-1)2n-1+n2n,两式相减可得-S=1+2+22+2n-1-n2n=(1-n)2n-1,则S=(n-1)2n+1,故Tn=(n-1)2n+2.13.已知首项为32的等比数列an不是递减数列,其前n项和为Sn(nN*),且S3+a3,S5+a5,S4+a4成等差数列.(1)求数列an的通项公式;(2)设bn=(-1)n+1n(nN*),求数列anbn的前n项和Tn.解:(1)设等比
14、数列an的公比为q.由S3+a3,S5+a5,S4+a4成等差数列,可得2(S5+a5)=S3+a3+S4+a4,即2(S3+a4+2a5)=2S3+a3+2a4,即4a5=a3,则q2=a5a3=14,解得q=12.由等比数列an不是递减数列,可得q=-12,故an=32-12n-1=(-1)n-132n.(2)由bn=(-1)n+1n,可得anbn=(-1)n-132n(-1)n+1n=3n12n.故前n项和Tn=3112+2122+n12n,则12Tn=31122+2123+n12n+1,两式相减可得,12Tn=312+122+12n-n12n+1=3121-12n1-12-n12n+1
15、,化简可得Tn=61-n+22n+1.14.若数列an的前n项和Sn满足Sn=2an-(0,nN*).(1)证明:数列an为等比数列,并求an;(2)若=4,bn=an,n是奇数,log2an,n是偶数(nN*),求数列bn的前2n项和T2n.答案:(1)证明Sn=2an-,当n=1时,得a1=,当n2时,Sn-1=2an-1-,则Sn-Sn-1=2an-2an-1,即an=2an-2an-1,an=2an-1,数列an是以为首项,2为公比的等比数列,an=2n-1.(2)解=4,an=42n-1=2n+1,bn=2n+1,n是奇数,n+1,n是偶数.T2n=22+3+24+5+26+7+22
16、n+2n+1=(22+24+26+22n)+(3+5+2n+1)=4-22n41-4+n(3+2n+1)2=4n+1-43+n(n+2),T2n=4n+13+n2+2n-43.高考预测15.在等差数列an中,公差d0,a10=19,且a1,a2,a5成等比数列.(1)求an;(2)设bn=an2n,求数列bn的前n项和Sn.解:(1)a1,a2,a5成等比数列,a22=a1a5,即(a1+d)2=a1(a1+4d).又a10=19=a1+9d,a1=1,d=2.an=2n-1.(2)bn=an2n=(2n-1)2n,Sn=2+322+(2n-1)2n.2Sn=22+323+(2n-3)2n+(2n-1)2n+1.由-,得-Sn=2+2(22+23+2n)-(2n-1)2n+1=222(2n-1-1)2-1+2-(2n-1)2n+1=(3-2n)2n+1-6.即Sn=(2n-3)2n+1+6.
