考点规范练25数列求和一、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+1-12n2.若数列{an}满足a1=1,且对任意的n∈N*都有an+1=a1+an+n,则{1an}的前100项和为()A.100101B.99100C.101100D.2001013.在数列{an}中,如果a1=1,a2=2,an+2-an=1+(-1)n,那么S100的值为()A.2500B.2600C.2700D.28004.已知函数f(x)=xa的图象过点(4,2),令an=1f(n+1)+f(n),n∈N*.记数列{an}的前n项和为Sn,则S2020等于()A.√2020-1B.√2020+1C.√2021-1D.√2021+15.已知数列{an}满足an+1+(-1)nan=2n-1,则{an}的前60项和为()A.3690B.3660C.1845D.18306.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+17.已知等差数列{an}满足a3=7,a5+a7=26,bn=1an2-1(n∈N*),数列{bn}的前n项和为Sn,则S100=.8.已知{an}是等差数列,{bn}是等比数列,且b2=3,b3=9,a1=b1,a14=b4.(1)求{an}的通项公式;(2)设cn=an+bn,求数列{cn}的前n项和.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d>1时,记cn=anbn,求数列{cn}的前n项和Tn.10.已知Sn为数列{an}的前n项和,an>0,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.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.二、能力提升12.几位大学生响应国家的创业号召,开发了一款应用软件.为激发大家学习数学的兴趣,他们推出了“解数学题获取软件激活码”的活动.这款软件的激活码为下面数学问题的答案:已知数列1,1,2,1,2,4,1,2,4,8,1,2,4,8,16,…,其中第一项是20,接下来的两项是20,21,再接下来的三项是20,21,22,依此类推.求满足如下条件的最小整数N:N>100且该数列的前N项和为2的整数幂.那么该款软件的激活码是()A.440B.330C.220D.11013.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(n∈N*),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+1·n(n∈N*),求数列{an·bn}的前n项和Tn.14.设数列{an}的前n项和为Sn,且Sn=2-an,n∈N*,设函数f(x)=log12x.数列{bn}满足bn=f(an),记{bn}的前n项和为Tn.(1)求an及Tn;(2)记cn=an·bn,求cn的最大值.三、高考预测15.已知等比数列{an}的各项均为正数,a1=1,公比为q;在等差数列{bn}中,b1=3,且{bn}的前n项和为Sn,a3+S3=27,q=S2a2.(1)求{an}与{bn}的通项公式;(2)设数列{cn}满足cn=32Sn,求{cn}的前n项和Tn.考点规范练25数列求和1.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.D解析 an+1=a1+an+n,∴an+1-an=1+n.∴an-an-1=n(n≥2).∴an=(an-an-1)+(an-1-an-2)+…+(a2-a1)+a1=n+(n-1)+…+2+1=n(n+1)2.∴1an=2n(n+1)=2(1n-1n+1).∴{1an}的前100项和为2¿1100-1101)=2(1-1101)=200101.故选D.3.B解析当n为奇数时,an+2-an=0,所以an=1,当n为偶数时,an+2-an=2,所以an=n,故an={1,n,为奇数n,n,为偶数于是S100=50+(2+100)×502=2600.4.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)=1√n+1+√n=√n+1−√n,S2020=a1+a2+a3+…+a2020=(√2−√1)+(√3−√2)+(√4−√3)+…+(√2021−√2020)=√2021-1.5.D解析 an+1+(-1)nan=2n-1,∴当n=2k(k∈N*)时,a2k+1+a2k=4k-1,①当n=2k+1(k∈N*)时,a2k+2-a2k+1=4k+1,②①+②得,a2k+a2k+2=8k.则a2+a4+a6+a8+…+a60=(a2+a4)+(a6+a8)+…+(a58+a60)=8×(1+3+…+29)=8×15×(1+29)2=1800.由②得a2k+1=a2k+2-(4k+1),∴a1+a3+a5+…+a59=a2+a4+…+a60-[4×(0+1+2+…+29)+30]=1800-(4×30×292+30)=30,∴a1+a2+…+a60=1800+30=1830.6.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)=12(12n-1-12n+1),∴Sn=12(11-13+13-15+…+12n-1-12n+1)=n2n+1,故选B.7.25101解析因为a3=7,a5+a7=26,所以公差d=2,...
