1、考点规范练32数列求和考点规范练B册第19页基础巩固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
2、=1+n.an-an-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
3、an=2n-70,解得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.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 020等于()A.2020-1B.2020+1C.2021-1D.2021+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,S2020=a1+a2+a3+a2020=(2-1)+(3
4、-2)+(4-3)+(2021-2020)=2021-1.5.(2019辽宁沈阳质量监测)已知数列an满足an+1+(-1)n+1an=2,则其前100项和为()A.250B.200C.150D.100答案:D解析:当n=2k-1时,a2k+a2k-1=2,an的前100项和为(a1+a2)+(a3+a4)+(a99+a100)=502=100,故选D.6.(2019广东深圳月考)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),
5、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.已知等差数列an,a5=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
6、-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.(2019新疆乌鲁木齐模拟)把函数f(x)=sin x(x0)所有的零点按从小到大的顺序排列,构成数列an,数列bn满足bn=3nan,则数列bn的前n项和Tn=.答案:(2n-1)3n+1+34解析:令sinx=0(x0),解得x=k,kN*,即an=n,nN*,则bn=3nan=n3n.则数列bn的前n项和Tn=(3+232+333+n3n),3Tn=32+233
7、+(n-1)3n+n3n+1,-,得-2Tn=(3+32+3n-n3n+1)=3(1-3n)1-3-n3n+1,可得Tn=(2n-1)3n+1+34.9.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=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+2
8、2n+1)=(-1)1-(-1)n1-(-1)+23(1-4n)1-4=(-1)n-12+83(4n-1).10.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=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-
9、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-1.11.(2019云南玉溪五调)若数列an的前n项和为Sn,首项a10,且2Sn=an2+an(nN*).(1)求数列an的通项公式;(2)若an0,令bn=4an(an+2),数列bn的前n项和为Tn.若Tn0,得an=n,bn=4n(n+2)=21n-1n+2,Tn=21-13+12-14+13-15+1n-1n+2=3-4n+6(n+1)(n+2)3
10、,又mZ,故mmin=3.12.(2019广东珠海高三上学期期末)已知Sn为等差数列an的前n项和,公差d=-2,且a1,a3,a4成等比数列.(1)求an,Sn;(2)设Tn=|a1|+|a2|+|an|,求Tn.解:(1)由a1,a3,a4成等比数列可知,a32=a1a4,从而(a1+2d)2=a1(a1+3d),又d=-2,解得a1=8.所以an=-2n+10.故Sn=na1+n(n-1)2d=-n2+9n.(2)由an=-2n+10,知当n0;当n=5时,an=0;当n5时,an5时,Tn=S5+|a6|+|an|=2S5-Sn=n2-9n+40.能力提升13.今要在一个圆周上标出一些
11、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+2.14.已知首项
12、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-132
13、n(-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,化简可得Tn=61-n+22n+1.15.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,log2an,n是偶数(nN*),求数列bn的前2n项和T2n.解:(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a
14、n=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)=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.高考预测16.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+6.8
