高中數(shù)學(xué)專題突破練習(xí)《等差數(shù)列前n項(xiàng)和及其性質(zhì)》含詳細(xì)答案解析

4.2.2等差數(shù)列的前n項(xiàng)和公式第1課時(shí)等差數(shù)列前n項(xiàng)和及其性質(zhì)基礎(chǔ)過關(guān)練題組一求等差數(shù)列的前n項(xiàng)和1.已知等差數(shù)列{an}滿足a1=1,am=99,d=2,則其前m項(xiàng)和Sm等于()A.2300 B.2400 C.2600 D.25002.在-20與40之間插入8個(gè)數(shù),使這10個(gè)數(shù)成等差數(shù)列,則這10個(gè)數(shù)的和為()A.200 B.100 C.90 D.703.設(shè)Sn是等差數(shù)列{an}的前n項(xiàng)和,已知a2=3,a6=11,則S7等于()A.13 B.35 C.49 D.634.(2020安徽合肥高三第一次教學(xué)質(zhì)量檢測(cè))已知等差數(shù)列{an}的前n項(xiàng)和為Sn,a1=-3,2a4+3a7=9,則S7等于()A.21 B.1 C.-42 D.05.若數(shù)列{an}為等差數(shù)列,Sn為其前n項(xiàng)和,且a1=2a5-1,則S17等于()A.-17 B.-172 C.1726.(2019湖南師大附中高二上期中)在等差數(shù)列{an}中,若a5,a7是方程x2-2x-6=0的兩個(gè)根,則數(shù)列{an}的前11項(xiàng)的和為()A.22 B.-33 C.-11 D.117.已知等差數(shù)列{an}.(1)若a6=10,a8=16,求S5;(2)若a2+a4=485,求S5題組二等差數(shù)列前n項(xiàng)和的性質(zhì)8.設(shè)等差數(shù)列{an}的前n項(xiàng)和為Sn,若S3=9,S6=36,則a7+a8+a9等于()A.63 B.45 C.36 D.279.在等差數(shù)列{an}中,Sn是其前n項(xiàng)和,且S2011=S2018,Sk=S2008,則正整數(shù)k為()A.2019 B.2020 C.2021 D.202210.含2n+1項(xiàng)的等差數(shù)列,其奇數(shù)項(xiàng)的和與偶數(shù)項(xiàng)的和之比為()A.2n+1n C.n-1n11.已知等差數(shù)列{an},{bn}的前n項(xiàng)和分別為Sn,Tn,若SnTn=3nA.87 B.4837 C.97題組三等差數(shù)列前n項(xiàng)和的應(yīng)用12.數(shù)列{an}為等差數(shù)列,它的前n項(xiàng)和為Sn,若Sn=(n+1)2+λ,則λ的值是()A.-2 B.-1 C.0 D.113.(2020山東濟(jì)南一中高二上期中)已知等差數(shù)列{an}的前9項(xiàng)和為27,a10=8,則a100=()A.100 B.99 C.98 D.9714.(2020山東青島高二上期末)已知數(shù)列{an}的前n項(xiàng)和為Sn,若an+1=an+2,S5=25,n∈N*,則a5=()A.7 B.5 C.9 D.315.(2020天津一中高二上期中)已知等差數(shù)列前3項(xiàng)的和為34,后3項(xiàng)的和為146,所有項(xiàng)的和為390,則這個(gè)數(shù)列的項(xiàng)數(shù)為()A.13 B.12 C.11 D.1016.若數(shù)列{an}的前n項(xiàng)和Sn=2n2-3n(n∈N*),則a1+a7等于()A.11 B.15 C.17 D.2217.(2019湖南懷化三中高二上期中)已知{an}是首項(xiàng)為a1,公差為d的等差數(shù)列,Sn是其前n項(xiàng)和,且S5=5,S6=-3.求數(shù)列{an}的通項(xiàng)公式及Sn.能力提升練題組一求等差數(shù)列的前n項(xiàng)和1.(2020湖南郴州高二上期中,)已知數(shù)列{an}是等差數(shù)列且an>0,設(shè)其前n項(xiàng)和為Sn.若a1+a9=a52,則S9A.36 B.18 C.27 D.92.(2020江西九江一中高二上期中,)等差數(shù)列{an}的前n項(xiàng)和為Sn,若a2+a7+a12=30,則S13等于()A.130 B.65 C.70 D.753.(2019湖北黃岡高一下期末,)如圖,將若干個(gè)點(diǎn)擺成三角形圖案,每條邊(包括兩個(gè)端點(diǎn))有n(n≥2,n∈N*)個(gè)點(diǎn),相應(yīng)的圖案中點(diǎn)的總數(shù)記為an,則a2+a3+a4+…+an等于()A.3n22C.3n(n4.(2020安徽阜陽高二上期末,)已知數(shù)列{an}中,a1=1,a2=2,對(duì)任意正整數(shù)n,an+2-an=2+cosnπ,Sn為{an}的前n項(xiàng)和,則S100=.

題組二等差數(shù)列前n項(xiàng)和的性質(zhì)5.()已知數(shù)列{an},{bn}均為等差數(shù)列,其前n項(xiàng)和分別記為An,Bn,滿足AnBn=4n+12n+3A.2117 B.3729 C.53296.()設(shè)等差數(shù)列{an}的前n項(xiàng)和為Sn,且Sm=-2,Sm+1=0,Sm+2=3,則m=.

7.(2019河北滄州一中高二期中,)在等差數(shù)列{an}中,前m(m為奇數(shù))項(xiàng)的和為135,其中偶數(shù)項(xiàng)之和為63,且am-a1=14,則a100=.

題組三等差數(shù)列前n項(xiàng)和的應(yīng)用8.(2020河北正定中學(xué)高二期末,)設(shè)Sn是等差數(shù)列{an}的前n項(xiàng)和,若a5a3=59,則A.1 B.-1 C.2 D.19.(2019陜西西安一中高二上月考,)設(shè)Sn(Sn≠0,n∈N*)是數(shù)列{an}的前n項(xiàng)和,且a1=-1,an+1=Sn·Sn+1,則Sn等于()A.n B.-nC.1n D.-10.()若數(shù)列{an}的前n項(xiàng)和Sn=n2-4n+2(n∈N*),則|a1|+|a2|+…+|a10|等于()A.15 B.35 C.66 D.10011.(2020天津耀華中學(xué)高二上期中,)數(shù)列{an}滿足an=1+2+3+…+nn(n∈N*),則數(shù)列1aA.nn+2 B.2nn+2 12.()已知數(shù)列{an}的前n項(xiàng)和Sn=n2+2n-1(n∈N*),則a1+a3+a5+…+a25=.

13.()已知等差數(shù)列的前三項(xiàng)依次為a,3,5a,前n項(xiàng)和為Sn,且Sk=121.(1)求a及k的值;(2)設(shè)數(shù)列{bn}的通項(xiàng)公式為bn=Snn,求{bn}的前n項(xiàng)和T14.()在數(shù)列{an}中,a1=8,a4=2,且滿足an+2-2an+1+an=0(n∈N*).(1)求數(shù)列{an}的通項(xiàng)公式;(2)設(shè)Tn=|a1|+|a2|+…+|an|,求Tn.深度解析

答案全解全析基礎(chǔ)過關(guān)練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設(shè)該等差數(shù)列為{an},其前n項(xiàng)和為Sn,則由題意可知,a1=-20,a10=40,所以S10=10×(-3.C由題意得,S7=7(a1+a4.D設(shè)等差數(shù)列{an}的公差為d,則2a4+3a7=2(-3+3d)+3(-3+6d)=9,解得d=1,∴S7=7a1+7×62×d=7×(-3)+7×3×1=0,故選5.D設(shè)等差數(shù)列{an}的公差為d,∵a1=2a5-1,∴a1=2(a1+4d)-1,∴a1+8d=1,即a9=1,∴S17=17×(a1+a6.D在等差數(shù)列{an}中,若a5,a7是方程x2-2x-6=0的兩個(gè)根,則a5+a7=2,∴a6=12(a5+a7∴數(shù)列{an}的前11項(xiàng)的和為11×(a1+a7.解析設(shè)等差數(shù)列{an}的首項(xiàng)為a1,公差為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由等差數(shù)列前n項(xiàng)和的性質(zhì)可知,S3,S6-S3,S9-S6構(gòu)成等差數(shù)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?yàn)榈炔顢?shù)列的前n項(xiàng)和Sn是關(guān)于n的二次函數(shù),所以由二次函數(shù)圖象的對(duì)稱性及S2011=S2018,Sk=S2008,可得2011+20182=2008+10.B設(shè)該等差數(shù)列為{an},其首項(xiàng)為a1,前n項(xiàng)和為Sn,則S奇=(n+1)(a1∵a1+a2n+1=a2+a2n,∴S奇S偶11.C由等差數(shù)列的性質(zhì)知a8b8=15(a1+a15)212.B∵等差數(shù)列前n項(xiàng)和Sn的形式為Sn=An2+Bn(A,B為常數(shù)),且Sn=(n+1)2+λ=n2+2n+1+λ,∴λ=-1.13.C設(shè)等差數(shù)列{an}的首項(xiàng)為a1,公差為d,由等差數(shù)列{an}的前9項(xiàng)和為27,a10=8,得9a1+9×82d=9a1+3614.C∵an+1=an+2,即an+1-an=2,∴{an}是公差為2的等差數(shù)列,設(shè)其首項(xiàng)為a1,則S5=5a1+5×42×2=25,解得a1∴a5=1+(5-1)×2=9.15.A設(shè)該等差數(shù)列為{an},其前n項(xiàng)和為Sn.由題意得,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,數(shù)列{an}為等差數(shù)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能力提升練1.B由a1+a9=a52得,2a5=a52,∴a5=2,∴S9=9(a1+a2.A解法一:設(shè)等差數(shù)列{an}的首項(xiàng)為a1,公差為d,則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,故選解法二:設(shè)等差數(shù)列{an}的首項(xiàng)為a1,∵a2+a7+a12=30,∴3a7=30,即a7=10,∴S13=13(a1+a13)3.C由題圖可知,a2=3,a3=6,a4=9,a5=12,依此類推,n每增加1,圖案中的點(diǎn)數(shù)增加3,所以相應(yīng)圖案中的點(diǎn)數(shù)構(gòu)成首項(xiàng)為a2=3,公差為3的等差數(shù)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解析當(dāng)n為奇數(shù)時(shí),an+2-an=1,即數(shù)列{an}的奇數(shù)項(xiàng)是以1為首項(xiàng),1為公差的等差數(shù)列;當(dāng)n為偶數(shù)時(shí),an+2-an=3,即數(shù)列{an}的偶數(shù)項(xiàng)是以2為首項(xiàng),3為公差的等差數(shù)列,所以S100=(a1+a3+…+a99)+(a2+a4+…+a100)=50×1+50×492+50×2+50×492×3=55.B由等差數(shù)列前n項(xiàng)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解題模板易錯(cuò)警示等差數(shù)列{an}的前n項(xiàng)和的表示形式為Sn=an2+bn(a,b為常數(shù)),解題時(shí)可采用這種形式簡(jiǎn)化運(yùn)算.本題要注意AnBn中有比例系數(shù)6.答案4解析因?yàn)镾n是等差數(shù)列{an}的前n項(xiàng)和,所以數(shù)列Snn是等差數(shù)列,所以Smm+Sm+2m+2=27.答案101解析設(shè)等差數(shù)列{an}的公差為d,前n項(xiàng)和為Sn,由題意可知,Sm=135,前m項(xiàng)中偶數(shù)項(xiàng)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,∴數(shù)列∴1Sn=-1+(n-1)×(-1)=-n,∴Sn=-1n10.C由Sn=n2-4n+2①得,當(dāng)n=1時(shí),a1=S1=1-4+2=-1,當(dāng)n≥2時(shí),Sn-1=(n-1)2-4(n-1)+2②,①-②得,an=2n-5(n≥2,n∈N*),經(jīng)檢驗(yàn),當(dāng)n=1時(shí),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.故選C.11.B依題意得,an=n(1+n∴1anan+1∴1a1a2+1=412-13+13-14+…+1n+1-1n12.答案350解析當(dāng)n=1時(shí),a1=S1=12+2×1-1=2;當(dāng)n≥2時(shí),an=Sn-Sn-1=2n+1,經(jīng)檢驗(yàn),當(dāng)n=1時(shí),不符合上式,∴an=2因此{(lán)an}除第1項(xiàng)外,其余項(xiàng)構(gòu)成以a2=5為首項(xiàng),2為公差的等差數(shù)列,從而a3,a5,…,a25是以a3=7為首項(xiàng),4為公差的等差數(shù)列,∴a1+a3+a5+…+a25=a1+12a13.解析(1)設(shè)該等差數(shù)列為{an},首項(xiàng)為a1,公差為d,則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(負(fù)值舍去)