高中数学专题突破练习《等差数列前n项和及其性质》含详细答案解析

4.2.2等差数列的前n项和公式第1课时等差数列前n项和及其性质基础过关练题组一求等差数列的前n项和1.已知等差数列{an}满足a1=1,am=99,d=2,则其前m项和Sm等于()A.2300 B.2400 C.2600 D.25002.在-20与40之间插入8个数,使这10个数成等差数列,则这10个数的和为()A.200 B.100 C.90 D.703.设Sn是等差数列{an}的前n项和,已知a2=3,a6=11,则S7等于()A.13 B.35 C.49 D.634.(2020安徽合肥高三第一次教学质量检测)已知等差数列{an}的前n项和为Sn,a1=-3,2a4+3a7=9,则S7等于()A.21 B.1 C.-42 D.05.若数列{an}为等差数列,Sn为其前n项和,且a1=2a5-1,则S17等于()A.-17 B.-172 C.1726.(2019湖南师大附中高二上期中)在等差数列{an}中,若a5,a7是方程x2-2x-6=0的两个根,则数列{an}的前11项的和为()A.22 B.-33 C.-11 D.117.已知等差数列{an}.(1)若a6=10,a8=16,求S5;(2)若a2+a4=485,求S5题组二等差数列前n项和的性质8.设等差数列{an}的前n项和为Sn,若S3=9,S6=36,则a7+a8+a9等于()A.63 B.45 C.36 D.279.在等差数列{an}中,Sn是其前n项和,且S2011=S2018,Sk=S2008,则正整数k为()A.2019 B.2020 C.2021 D.202210.含2n+1项的等差数列,其奇数项的和与偶数项的和之比为()A.2n+1n C.n-1n11.已知等差数列{an},{bn}的前n项和分别为Sn,Tn,若SnTn=3nA.87 B.4837 C.97题组三等差数列前n项和的应用12.数列{an}为等差数列,它的前n项和为Sn,若Sn=(n+1)2+λ,则λ的值是()A.-2 B.-1 C.0 D.113.(2020山东济南一中高二上期中)已知等差数列{an}的前9项和为27,a10=8,则a100=()A.100 B.99 C.98 D.9714.(2020山东青岛高二上期末)已知数列{an}的前n项和为Sn,若an+1=an+2,S5=25,n∈N*,则a5=()A.7 B.5 C.9 D.315.(2020天津一中高二上期中)已知等差数列前3项的和为34,后3项的和为146,所有项的和为390,则这个数列的项数为()A.13 B.12 C.11 D.1016.若数列{an}的前n项和Sn=2n2-3n(n∈N*),则a1+a7等于()A.11 B.15 C.17 D.2217.(2019湖南怀化三中高二上期中)已知{an}是首项为a1,公差为d的等差数列,Sn是其前n项和,且S5=5,S6=-3.求数列{an}的通项公式及Sn.能力提升练题组一求等差数列的前n项和1.(2020湖南郴州高二上期中,)已知数列{an}是等差数列且an>0,设其前n项和为Sn.若a1+a9=a52,则S9A.36 B.18 C.27 D.92.(2020江西九江一中高二上期中,)等差数列{an}的前n项和为Sn,若a2+a7+a12=30,则S13等于()A.130 B.65 C.70 D.753.(2019湖北黄冈高一下期末,)如图,将若干个点摆成三角形图案,每条边(包括两个端点)有n(n≥2,n∈N*)个点,相应的图案中点的总数记为an,则a2+a3+a4+…+an等于()A.3n22C.3n(n4.(2020安徽阜阳高二上期末,)已知数列{an}中,a1=1,a2=2,对任意正整数n,an+2-an=2+cosnπ,Sn为{an}的前n项和,则S100=.

题组二等差数列前n项和的性质5.()已知数列{an},{bn}均为等差数列,其前n项和分别记为An,Bn,满足AnBn=4n+12n+3A.2117 B.3729 C.53296.()设等差数列{an}的前n项和为Sn,且Sm=-2,Sm+1=0,Sm+2=3,则m=.

7.(2019河北沧州一中高二期中,)在等差数列{an}中,前m(m为奇数)项的和为135,其中偶数项之和为63,且am-a1=14,则a100=.

题组三等差数列前n项和的应用8.(2020河北正定中学高二期末,)设Sn是等差数列{an}的前n项和,若a5a3=59,则A.1 B.-1 C.2 D.19.(2019陕西西安一中高二上月考,)设Sn(Sn≠0,n∈N*)是数列{an}的前n项和,且a1=-1,an+1=Sn·Sn+1,则Sn等于()A.n B.-nC.1n D.-10.()若数列{an}的前n项和Sn=n2-4n+2(n∈N*),则|a1|+|a2|+…+|a10|等于()A.15 B.35 C.66 D.10011.(2020天津耀华中学高二上期中,)数列{an}满足an=1+2+3+…+nn(n∈N*),则数列1aA.nn+2 B.2nn+2 12.()已知数列{an}的前n项和Sn=n2+2n-1(n∈N*),则a1+a3+a5+…+a25=.

13.()已知等差数列的前三项依次为a,3,5a,前n项和为Sn,且Sk=121.(1)求a及k的值;(2)设数列{bn}的通项公式为bn=Snn,求{bn}的前n项和T14.()在数列{an}中,a1=8,a4=2,且满足an+2-2an+1+an=0(n∈N*).(1)求数列{an}的通项公式;(2)设Tn=|a1|+|a2|+…+|an|,求Tn.深度解析

答案全解全析基础过关练1.D解法一:由am=a1+(m-1)d,得99=1+(m-1)×2,解得m=50,所以Sm=S50=50×1+50×492×2=2解法二:同解法一,得m=50,所以Sm=S50=50(a1+a50)2.B设该等差数列为{an},其前n项和为Sn,则由题意可知,a1=-20,a10=40,所以S10=10×(-3.C由题意得,S7=7(a1+a4.D设等差数列{an}的公差为d,则2a4+3a7=2(-3+3d)+3(-3+6d)=9,解得d=1,∴S7=7a1+7×62×d=7×(-3)+7×3×1=0,故选5.D设等差数列{an}的公差为d,∵a1=2a5-1,∴a1=2(a1+4d)-1,∴a1+8d=1,即a9=1,∴S17=17×(a1+a6.D在等差数列{an}中,若a5,a7是方程x2-2x-6=0的两个根,则a5+a7=2,∴a6=12(a5+a7∴数列{an}的前11项的和为11×(a1+a7.解析设等差数列{an}的首项为a1,公差为d.(1)∵a6=10,a8=16,∴a1+5∴S5=5a1+5×42(2)解法一:∵a2+a4=a1+d+a1+3d=485∴a1+2d=245∴S5=5a1+5×42d=5a1+10d=5(a1+2d)=5×24解法二:∵a2+a4=a1+a5,∴a1+a5=485∴S5=5(a1+a8.B由等差数列前n项和的性质可知,S3,S6-S3,S9-S6构成等差数列,所以S3+(S9-S6)=2(S6-S3),即S9=3S6-3S3,又S3=9,S6=36,所以S9=3×36-3×9=81,所以a7+a8+a9=S9-S6=81-36=45.9.C因为等差数列的前n项和Sn是关于n的二次函数,所以由二次函数图象的对称性及S2011=S2018,Sk=S2008,可得2011+20182=2008+10.B设该等差数列为{an},其首项为a1,前n项和为Sn,则S奇=(n+1)(a1∵a1+a2n+1=a2+a2n,∴S奇S偶11.C由等差数列的性质知a8b8=15(a1+a15)212.B∵等差数列前n项和Sn的形式为Sn=An2+Bn(A,B为常数),且Sn=(n+1)2+λ=n2+2n+1+λ,∴λ=-1.13.C设等差数列{an}的首项为a1,公差为d,由等差数列{an}的前9项和为27,a10=8,得9a1+9×82d=9a1+3614.C∵an+1=an+2,即an+1-an=2,∴{an}是公差为2的等差数列,设其首项为a1,则S5=5a1+5×42×2=25,解得a1∴a5=1+(5-1)×2=9.15.A设该等差数列为{an},其前n项和为Sn.由题意得,a1+a2+a3=34,an-2+an-1+an=146,∴(a1+a2+a3)+(an-2+an-1+an)=(a1+an)+(a2+an-1)+(a3+an-2)=3(a1+an)=34+146,∴a1+an=60.又Sn=n(a1+an)2,16.D由Sn=2n2-3n(n∈N*)可知,数列{an}为等差数列,所以S7=7×(a1+a7)2=2×72-3×7,17.解析由S5=5,S6=-3,得5a1+∴an=7+(n-1)×(-3)=-3n+10(n∈N*),Sn=n[7+(-3n+10)]2=-32能力提升练1.B由a1+a9=a52得,2a5=a52,∴a5=2,∴S9=9(a1+a2.A解法一:设等差数列{an}的首项为a1,公差为d,则a2+a7+a12=(a1+d)+(a1+6d)+(a1+11d)=3a1+18d=30,∴a1+6d=10.∴S13=13a1+13×122d=13(a1+6d)=13×10=130,故选解法二:设等差数列{an}的首项为a1,∵a2+a7+a12=30,∴3a7=30,即a7=10,∴S13=13(a1+a13)3.C由题图可知,a2=3,a3=6,a4=9,a5=12,依此类推,n每增加1,图案中的点数增加3,所以相应图案中的点数构成首项为a2=3,公差为3的等差数列,所以an=3+(n-2)×3=3n-3,n≥2,n∈N*,所以a2+a3+a4+…+an=(n-1)(3+34.答案5050解析当n为奇数时,an+2-an=1,即数列{an}的奇数项是以1为首项,1为公差的等差数列;当n为偶数时,an+2-an=3,即数列{an}的偶数项是以2为首项,3为公差的等差数列,所以S100=(a1+a3+…+a99)+(a2+a4+…+a100)=50×1+50×492+50×2+50×492×3=55.B由等差数列前n项和的特征及AnBn=4n+12n∴a5=A5-A4=5×(4×5+1)k-4×(4×4+1)k=37k,b7=B7-B6=7×(2×7+3)k-6×(2×6+3)k=29k.∴a5b7=37k29解题模板易错警示等差数列{an}的前n项和的表示形式为Sn=an2+bn(a,b为常数),解题时可采用这种形式简化运算.本题要注意AnBn中有比例系数6.答案4解析因为Sn是等差数列{an}的前n项和,所以数列Snn是等差数列,所以Smm+Sm+2m+2=27.答案101解析设等差数列{an}的公差为d,前n项和为Sn,由题意可知,Sm=135,前m项中偶数项之和S偶=63,∴S奇=135-63=72,∴S奇-S偶=a1+(m-1)d∵Sm=m(a1又∵am-a1=14,am=a1+(m-1)d,∴a1=2,d=am-a∴a100=a1+99d=101.8.AS9S5=92(a1+a9)9.D∵an+1=Sn+1-Sn,∴Sn+1-Sn=Sn+1·Sn,又∵Sn≠0,∴1Sn+1-1Sn=-1.又S1=a1=-1,∴1S1=-1,∴数列∴1Sn=-1+(n-1)×(-1)=-n,∴Sn=-1n10.C由Sn=n2-4n+2①得,当n=1时,a1=S1=1-4+2=-1,当n≥2时,Sn-1=(n-1)2-4(n-1)+2②,①-②得,an=2n-5(n≥2,n∈N*),经检验,当n=1时,不符合an=2n-5,∴an=-1,n=1,2n-5,n≥2,n∴n≥3.∴|a1|+|a2|+…+|a10|=1+1+a3+…+a10=2+(S10-S2)=2+[(102-4×10+2)-(22-4×2+2)]=66.故选C.11.B依题意得,an=n(1+n∴1anan+1∴1a1a2+1=412-13+13-14+…+1n+1-1n12.答案350解析当n=1时,a1=S1=12+2×1-1=2;当n≥2时,an=Sn-Sn-1=2n+1,经检验,当n=1时,不符合上式,∴an=2因此{an}除第1项外,其余项构成以a2=5为首项,2为公差的等差数列,从而a3,a5,…,a25是以a3=7为首项,4为公差的等差数列,∴a1+a3+a5+…+a25=a1+12a13.解析(1)设该等差数列为{an},首项为a1,公差为d,则a1=a,a2=3,a3=5a.由已知得a+5a=6,得a=1,∴a1=1,a2=3,a3=5,∴d=2,∴Sk=ka1+k(k-1)2由Sk=k2=121,得k=11(负值舍去)