新教材2024-2025學(xué)年高中數(shù)學(xué)第一章數(shù)列2等差數(shù)列2.2等差數(shù)列的前n項(xiàng)和第1課時(shí)等差數(shù)列前n項(xiàng)和的推導(dǎo)及初步應(yīng)用分層作業(yè)北師大版選擇性必修第二冊(cè)

第1課時(shí)等差數(shù)列前n項(xiàng)和的推導(dǎo)及初步應(yīng)用A級(jí)必備學(xué)問(wèn)基礎(chǔ)練1.已知等差數(shù)列{an}滿意a1=1,am=99,d=2,則其前m項(xiàng)和Sm等于()A.2300 B.2400 C.2600 D.25002.已知等差數(shù)列{an}的前n項(xiàng)和是Sn,a3=1,S7=3a6,則S3=()A.1 B.-1 C.3 D.-33.[2024云南曲靖民族中學(xué)?？计谥衇在等差數(shù)列{an}中,已知a4+a8=16,則該數(shù)列前11項(xiàng)的和為()A.44 B.88 C.99 D.1104.記等差數(shù)列的前n項(xiàng)和為Sn,若S2=4,S4=20,則該數(shù)列的公差d等于()A.2 B.3 C.6 D.75.已知在數(shù)列{an}中,a1=1,an=an-1+12(n≥2,n∈N+),則數(shù)列{an}的前9項(xiàng)和等于(A.27 B.632 C.45 D.-6.設(shè)Sn為等差數(shù)列{an}的前n項(xiàng)和,若S3=3,S6=24,則a9=.

7.已知等差數(shù)列{an}的通項(xiàng)公式為an=2n-1(n∈N+),那么它的前n項(xiàng)和Sn=.

8.已知等差數(shù)列{an}的前n項(xiàng)和為Sn,且a3+a4+a5+a6+a7=150,則S9=.

9.[2024河南洛陽(yáng)第三中學(xué)校聯(lián)考模擬預(yù)料]已知數(shù)列{an},等差數(shù)列{bn}滿意bn=an+an+1,b1=-3,b2+b3=-12.(1)證明:an-an+2=2;(2)若{an}為等差數(shù)列,求數(shù)列{an}的前n項(xiàng)和.B級(jí)關(guān)鍵實(shí)力提升練10.(多選題)[2024山東日照高二統(tǒng)考期中]已知{an}是等差數(shù)列,其公差為d,前n項(xiàng)和為Sn,a10=10,S10=70,則()A.a1=4 B.d=2C.數(shù)列{an}為遞減數(shù)列 D.數(shù)列Sn11.在等差數(shù)列{an}中,a32+a82+2a3a8=9,且an<A.-9 B.-11 C.-13 D.-1512.在等差數(shù)列{an}和{bn}中,a1=25,b1=75,a100+b100=100,則數(shù)列{an+bn}的前100項(xiàng)的和為()A.10000 B.8000 C.9000 D.1100013.(多選題)[2024安徽安慶一中?？茧A段練習(xí)]已知等差數(shù)列{an}的前n項(xiàng)和為Sn,若S11=a5+aA.S11=0 B.a6=0 C.S6=S5 D.S7=S614.(多選題)[2024貴州黔東南高二階段練習(xí)]設(shè)等差數(shù)列{an}的前n項(xiàng)和為Sn.若S3=0,a4=4,則()A.an=4n-8 B.an=2n-4C.Sn=2n2-6n D.Sn=n2-3n15.正整數(shù)數(shù)列前n(n∈N+)個(gè)奇數(shù)的和為.

16.已知等差數(shù)列{an}的前n項(xiàng)和為Sn,若OB=a1OA+a200OC,且A,B,C三點(diǎn)共線(該直線不過(guò)原點(diǎn)O),則S200=.

17.記Sn是公差不為0的等差數(shù)列{an}的前n項(xiàng)和,若a3=S5,a1a4=a5,則an=.

18.已知公差大于零的等差數(shù)列{an}的前n項(xiàng)和為Sn,且滿意:a3a4=117,a2+a5=22.(1)求數(shù)列{an}的通項(xiàng)公式an;(2)若數(shù)列{bn}是等差數(shù)列,且bn=Snn19.[2024遼寧朝陽(yáng)高二階段練習(xí)]已知等差數(shù)列{an}的前n項(xiàng)和為Sn,a4=13,S7=13a1.(1)求數(shù)列{an}的通項(xiàng)公式;(2)求證:{Sn+9C級(jí)學(xué)科素養(yǎng)創(chuàng)新練20.已知數(shù)列{an}的全部項(xiàng)均為正數(shù),其前n項(xiàng)和為Sn,且Sn=14an2+12an-3(1)證明:{an}是等差數(shù)列;(2)求數(shù)列{an}的通項(xiàng)公式.

參考答案2.2等差數(shù)列的前n項(xiàng)和第1課時(shí)等差數(shù)列前n項(xiàng)和的推導(dǎo)及初步應(yīng)用1.D由am=a1+(m-1)d,得99=1+(m-1)×2,解得m=50,所以S50=50×1+50×492×22.D由已知設(shè)等差數(shù)列的公差為d,則a3=a1+2d=1,7a1+7×62d=3(a1+解得a1=-3,d=2,所以S3=3a1+3d=-3.故選D.3.B由等差數(shù)列的性質(zhì)可知,a1+a11=a4+a8=16,故前11項(xiàng)的和為S11=11(a1故選B.4.B(方法一)由S2=2a1(方法二)由S4-S2=a3+a4=a1+2d+a2+2d=S2+4d,所以20-4=4+4d,解得d=3.5.A由已知數(shù)列{an}是以1為首項(xiàng),以12為公差的等差數(shù)列,∴S9=9×1+9×82×126.15設(shè)等差數(shù)列的公差為d,則S3=3a1+3×22d=3a1+3d=3,即aS6=6a1+6×52d=6a1+15d=24,即2a1+5由a1+故a9=a1+8d=-1+8×2=15.7.n2∵等差數(shù)列{an}的通項(xiàng)公式為an=2n-1(n∈N+),∴Sn=n(1+2n8.270因?yàn)樵诘炔顢?shù)列{an}中,a3+a4+a5+a6+a7=5a5=150,a5=30,則S9=9(a1+a9)9.(1)證明設(shè)數(shù)列{bn}的公差為d,由b2+b3=-12,得2b1+3d=-12,由b1=-3,解得d=-2,故bn=-2n-1,即an+an+1=-2n-1, ①則an+1+an+2=-2n-3, ②①-②得an-an+2=2,故an-an+2=2得證.(2)解若{an}為等差數(shù)列,設(shè)公差為d',由an+2-an=-2,可得2d'=-2,則d'=-1.又an+an+1=-2n-1,∴2an+d'=-2n-1,∴an=-n,∴a1=-1,∴{an}的前n項(xiàng)和Sn=(-1-n10.AB由題意可得a10=a1+9dSn-Sn-1=4+23(n-1)=23n+103(n≥2)不是常數(shù),故數(shù)列{S故選AB.11.D由a32+a82+2a3a8=9,得(a3∵an<0,∴a3+a8=-3,∴S10=10(a112.A由已知得{an+bn}為等差數(shù)列,故其前100項(xiàng)的和為S100=100[(a1+b1)+(13.ABC因?yàn)閧an}是等差數(shù)列,所以a5+a7=a1+a11=2a6.依據(jù)題意S11=a5+a又因?yàn)镾11=11(a1+a11)2=11從而a6=0,S11=0,故選項(xiàng)A,B正確;又因?yàn)閍6=S6-S5=0,即S6=S5,故選項(xiàng)C正確;對(duì)于選項(xiàng)D,S7-S6=a7,依據(jù)題意無(wú)法推斷a7是否為零,故選項(xiàng)D錯(cuò)誤.故選ABC.14.BD由題意,S3=3所以an=-2+(n-1)·2=2n-4,Sn=-2n+2n(n-1故選BD.15.n21+3+5+…+(2n-1)=n(1+2n16.100因?yàn)锳,B,C三點(diǎn)共線(該直線不過(guò)原點(diǎn)O),所以a1+a200=1,所以S200=200(a117.3-n設(shè)等差數(shù)列{an}的公差為d,∵a3=S5,a1a4=a5,∴a1+2d=5a1+5×42d,a1(a1+3d)=a1+解得a1=2,d=-1,則an=2-(n-1)=3-n.18.解(1)設(shè)等差數(shù)列{an}的公差為d,且d>0.∵a3+a4=a2+a5=22,又a3a4=117,∴a3,a4是方程x2-22x+117=0的兩個(gè)根.又公差d>0,∴a3<a4,∴a3=9,a4=13.∴a1+2d=9,a1+3d=13,∴(2)由(1)知,Sn=n×1+n(n-1)2×4∴bn=Snn+c=2n2-nn+c.∴b1∵{bn}是等差數(shù)列,∴2b2=b1+b3,∴2c2+c=0,∴c=-12(c=0舍去)經(jīng)檢驗(yàn),c=-12符合題意,∴c=-119.(1)解設(shè)等差數(shù)列{an}的公差為d,由a4=13,S7=13a1,得a1+3d=13,7a1+7×62d=13a1,解得a1=7,d=2,所以an=a1+(n-1)d=2(2)證明結(jié)合(1)可得Sn=na1+n(n-1)2d=7n+n2-n=n2+故S1+9=4,Sn+1+9-Sn+9=所以數(shù)列Sn+920.(1)證明當(dāng)n=1時(shí),a1=S1=14a12+解得a1=3或