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.已知数列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-
2、1=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
3、n72.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.(2020新疆二模)已知等差数列an满足a1=-2 018,其前n项和为Sn,若5S12-6S10=120,则S2 020=()A.-4 040B.-2 020C.2 020D.4 040答案:C解析:由5S12-6S10=120,得512a1+12112d-610a1+1092d=120,化简得60d=120,解得d=2.又已知a1=-2018,则S2020=2020(-2018)+2020201922=202
4、0.5.(2020江西南昌模拟)已知数列an为等差数列,Sn是其前n项和,a2=5,S5=35.数列1anan+1的前n项和为Tn,若对一切nN*都有2m+1Tn恒成立,则m能取到的最小整数为()A.-1B.0C.1D.2答案:B解析:设等差数列an的首项为a1,公差为d.由题意得a1+d=5,5a1+542d=35,解得a1=3,d=2,故an=3+2(n-1)=2n+1.所以1anan+1=1(2n+1)(2n+3)=1212n+1-12n+3,所以Tn=1213-15+15-17+12n+1-12n+3=1213-12n+3Tn恒成立,只需满足2m+116即可,故m能取到的最小整数为0.
5、6.已知数列an的前n项和为Sn,且满足a1=1,a2=2,Sn+1=an+2-an+1(nN*),则Sn=.答案:2n-1解析:Sn+1=an+2-an+1(nN*),Sn+1=Sn+2-Sn+1-(Sn+1-Sn),则Sn+2+1=2(Sn+1+1).由a1=1,a2=2,可得S2+1=2(S1+1),Sn+1+1=2(Sn+1)对任意的nN*都成立,数列Sn+1是首项为2,公比为2的等比数列,Sn+1=2n,即Sn=2n-1.7.已知数列an满足:a3=15,an-an+1=2anan+1,则数列anan+1前10项的和为.答案:1021解析:an-an+1=2anan+1,an-an+
6、1anan+1=2,即1an+1-1an=2.数列1an是以2为公差的等差数列.1a3=5,1an=5+2(n-3)=2n-1.an=12n-1.anan+1=1(2n-1)(2n+1)=1212n-1-12n+1.数列anan+1前10项的和为121-13+13-15+1210-1-1210+1=121-121=122021=1021.8.在数列an中,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=2
7、n+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=
8、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-12n=3-2n+32n,故Tn=6-2n+32n-1.10.已知数列an是各项均为正数的等比数列,且a1+a2=21a1+1a2,a3+a4=321a3+1a4.(1)求数列an的通项公式;(2)设bn=an2+log2a
9、n,求数列bn的前n项和Tn.解：(1)设等比数列an的公比为q(q0),则an=a1qn-1,且an0.由已知得a1+a2=21a1+1a2=2(a1+a2)a1a2,即a1a2=2;a3+a4=321a3+1a4=32(a3+a4)a3a4,即a3a4=32.a3a4=a1a2q4=2q4=32,且q0,q=2.由a1a2=a12q=2,且a10,得a1=1.数列an的通项公式为an=2n-1.(2)由(1)知bn=an2+log2an=4n-1+n-1,Tn=(1+4+42+4n-1)+(0+1+2+3+n-1)=1-4n1-4+n(n-1)2=4n-13+n(n-1)2.11.(202
10、0广西南宁三中模拟)设等差数列an的前n项和为Sn,且满足a3=3,S4=10.(1)求an的通项公式;(2)设bn=1Sn,求数列bn的前n项和Tn.解：(1)设等差数列an的公差为d,则a3=a1+2d=3,S4=4a1+432d=10,整理得a1+2d=3,2a1+3d=5,解得a1=1,d=1.等差数列an的通项公式为an=1+(n-1)1=n.(2)由(1)知,Sn=n(n+1)2,bn=1Sn=2n(n+1)=21n-1n+1.Tn=b1+b2+bn=21-12+212-13+21n-1n+1=21-12+12-13+1n-1n+1=21-1n+1=2nn+1.能力提升12.在数列
11、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.13.(2020黑龙江大庆模拟)数列1,x,1,x,x,1,x,x,x,1,x,x,x,x,1,x,其中在第n个1与第n+1个1之间插入n个x,
12、若该数列的前2 020项的和为7 891,则x=.答案:4解析:当n2时,前n个1之间共有n+1+2+3+(n-1)=n(n+1)2项,当n=63时,有63642=2016项,在第63个1的后面再跟的第4个x就是第2020项,所以前2020项中含63个1,其余的均为x.故该数列前2020项的和为631+(2020-63)x=7891,解得x=4.14.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)
13、因为an+12=2Sn+n+4,所以an2=2Sn-1+n-1+4(n2).两式相减,得an+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.由a22=2S1+1+4,可得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T
14、n=c1+c2+cn=-1+2-3+(-1)nn-123+134+1(n+1)(n+2).设Fn=-1+2-3+(-1)nn,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.15.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,
15、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+22n+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
16、预测16.正项等差数列an满足a1=4,且a2,a4+2,2a7-8成等比数列,an的前n项和为Sn.(1)求数列an的通项公式;(2)令bn=1Sn+2,求数列bn的前n项和Tn.解：(1)设数列an的公差为d(d0).由已知得a2(2a7-8)=(a4+2)2,且a1=4,化简得,d2+4d-12=0,解得d=2或d=-6(舍),所以an=a1+(n-1)d=2n+2.(2)因为Sn=n(a1+an)2=n(2n+6)2=n2+3n,所以bn=1Sn+2=1n2+3n+2=1(n+1)(n+2)=1n+1-1n+2,所以Tn=b1+b2+b3+bn=12-13+13-14+14-15+1n+1-1n+2=12-1n+2=n2n+4.