【全程復(fù)習(xí)方略】2020年數(shù)學(xué)文(廣西用)課時(shí)作業(yè)：第三章-第四節(jié)數(shù)列求和

溫馨提示：此套題為Word版，請按住Ctrl,滑動鼠標(biāo)滾軸，調(diào)整合適的觀看比例，答案解析附后。關(guān)閉Word文檔返回原板塊。課時(shí)提升作業(yè)(十六)一、選擇題1.(2021·崇左模擬)若數(shù)列{an}滿足:a1=19,an+1=an-3(n∈N*),則數(shù)列{an}的前n項(xiàng)和數(shù)值最大時(shí),n的值是()(A)6 (B)7 (C)8 (D)92.數(shù)列{an}的前n項(xiàng)和為Sn,若an=QUOTE,則S10等于()(A)QUOTE (B)QUOTE (C)QUOTE (D)QUOTE3.(2021·長春模擬)在等差數(shù)列{an}中,a9=QUOTEa12+6,則數(shù)列{an}的前11項(xiàng)和S11等于()(A)24 (B)48 (C)66 (D)1324.已知數(shù)列{an}的通項(xiàng)公式是an=2n-3(QUOTE)n,則其前20項(xiàng)和為()(A)380-QUOTE(1-QUOTE) (B)400-QUOTE(1-QUOTE)(C)420-QUOTE(1-QUOTE) (D)440-QUOTE(1-QUOTE)5.(2021·太原模擬)已知等差數(shù)列{an}的公差d≠0,且a1,a3,a9成等比數(shù)列,則QUOTE=()(A)QUOTE (B)QUOTE (C)QUOTE (D)QUOTE6.數(shù)列{an}的前n項(xiàng)和Sn=3n+b(b是常數(shù)),若這個(gè)數(shù)列是等比數(shù)列,那么b為()(A)3 (B)0 (C)-1 (D)17.等差數(shù)列{an}的前n項(xiàng)和為Sn,已知am-1+am+1-QUOTE=0,S2m-1=38,則m=()(A)38 (B)20 (C)10 (D)98.(力氣挑戰(zhàn)題)數(shù)列{an}的前n項(xiàng)和Sn=2n-1,則QUOTE+QUOTE+QUOTE+…+QUOTE等于()(A)(2n-1)2 (B)QUOTE(2n-1)2(C)4n-1 (D)QUOTE(4n-1)二、填空題9.已知等差數(shù)列{an}的前n項(xiàng)和為Sn.若a3=20-a6,則S8等于.10.數(shù)列{1+2n-1}的前n項(xiàng)和為.11.(2021·南寧模擬)已知數(shù)列{an}(n∈N*),其前n項(xiàng)和為Sn,給出下列四個(gè)命題:①若{an}是等差數(shù)列,則三點(diǎn)(10,QUOTE),(100,QUOTE),(110,QUOTE)共線;②若{an}是等差數(shù)列,且a1=-11,a3+a7=-6,則S1,S2,…,Sn這n個(gè)數(shù)中必定存在一個(gè)最大值;③若{an}是等比數(shù)列,則Sm,S2m-Sm,S3m-S2m(m∈N*)也是等比數(shù)列;④若Sn+1=a1+qSn(其中常數(shù)a1,q≠0),則{an}是等比數(shù)列.其中正確命題的序號是(將你認(rèn)為的正確命題的序號都填上).12.(2021·哈爾濱模擬)在數(shù)列{an}中,若對任意的n均有an+an+1+an+2為定值(n∈N*),且a7=2,a9=3,a98=4,則此數(shù)列{an}的前100項(xiàng)的和S100=.三、解答題13.已知數(shù)列{log2(an-1)}(n∈N*)為等差數(shù)列,且a1=3,a3=9.(1)求數(shù)列{an}的通項(xiàng)公式.(2)求和:Sn=QUOTE+QUOTE+…+QUOTE.14.等差數(shù)列{an}中,2a1+3a2=11,2a3=a2+a6-4,其前n項(xiàng)和為Sn.(1)求數(shù)列{an}的通項(xiàng)公式.(2)設(shè)數(shù)列{bn}滿足bn=QUOTE,其前n項(xiàng)和為Tn,求證:Tn<QUOTE(n∈N*).15.(2021·百色模擬)已知等差數(shù)列{an}中,a3+a5=10,{an}的前n項(xiàng)和為Sn,S5=15.(1)求數(shù)列{an}的通項(xiàng)公式an.(2)設(shè)bn=(QUOTE)n·an,求數(shù)列{bn}的前n項(xiàng)和Tn.16.(力氣挑戰(zhàn)題)已知數(shù)列{an}的通項(xiàng)公式是an=n·2n-1,bn=QUOTE,求數(shù)列{bn}的前n項(xiàng)和.答案解析1.【解析】選B.由已知a1=19,an+1-an=-3可得數(shù)列{an}是以19為首項(xiàng),以-3為公差的等差數(shù)列.方法一:故Sn=19n+QUOTE×(-3)=-QUOTEn2+QUOTEn=-QUOTE(n-QUOTE)2+QUOTE.所以當(dāng)n=7時(shí),Sn最大.方法二:故an=19+(n-1)×(-3)=-3n+22.令an≥0得-3n+22≥0,n≤QUOTE.又∵n∈N*,故n=7.2.【解析】選D.an=QUOTE=QUOTE(QUOTE-QUOTE),所以S10=a1+a2+…+a10=QUOTE(1-QUOTE+QUOTE-QUOTE+…+QUOTE-QUOTE)=QUOTE(1+QUOTE-QUOTE-QUOTE)=QUOTE,故選D.3.【解析】選D.設(shè)公差為d,則a1+8d=QUOTEa1+QUOTEd+6,即a1+5d=12,即a6=12,所以S11=11a6=132.4.【解析】選C.由an=2n-3(QUOTE)n,得S20=2(1+2+…+20)-3(QUOTE+QUOTE+…+QUOTE)=2×QUOTE-3×QUOTE=420-QUOTE(1-QUOTE),故選C.5.【解析】選C.等差數(shù)列{an}中,a1=a1,a3=a1+2d,a9=a1+8d,由于a1,a3,a9恰好構(gòu)成等比數(shù)列,所以有QUOTE=a1a9,即(a1+2d)2=a1(a1+8d),解得d=a1,所以該等差數(shù)列的通項(xiàng)為an=nd.則QUOTE的值為QUOTE.6.【思路點(diǎn)撥】依據(jù)數(shù)列的前n項(xiàng)和減去前n-1項(xiàng)的和得到數(shù)列的第n項(xiàng)的通項(xiàng)公式,即可得到此等比數(shù)列的首項(xiàng)與公比,依據(jù)首項(xiàng)和公比,利用等比數(shù)列的前n項(xiàng)和公式表示出前n項(xiàng)的和,與已知的Sn=3n+b對比后,即可得到b的值.【解析】選C.由于an=Sn-Sn-1=(3n+b)-(3n-1+b)=3n-3n-1=2×3n-1(n≥2),所以此數(shù)列是首項(xiàng)為2,公比為3的等比數(shù)列,則Sn=QUOTE=3n-1,所以b=-1.7.【解析】選C.由于{an}是等差數(shù)列,所以am-1+am+1=2am,由am-1+am+1-QUOTE=0,得2am-QUOTE=0,所以am=2(am=0舍),又S2m-1=38,即QUOTE=38,即(2m-1)×2=38,解得m=10,故選C.8.【解析】選D.an=Sn-Sn-1=2n-1(n>1),又a1=S1=1=20,適合上式,∴an=2n-1(n∈N*),∴{QUOTE}是QUOTE=1,q=22的等比數(shù)列,由求和公式得QUOTE+QUOTE+QUOTE+…+QUOTE=QUOTE=QUOTE(4n-1).9.【解析】由于a3=20-a6,所以S8=4(a3+a6)=4×20=80.答案:8010.【解析】前n項(xiàng)和Sn=(1+20)+(1+21)+(1+22)+…+(1+2n-1)=n+QUOTE=n+2n-1.答案:n+2n-111.【解析】①中設(shè)數(shù)列{an}的公差為d,由于A(10,QUOTE),B(100,QUOTE),C(110,QUOTE).則kAB=QUOTE=QUOTE,kBC=QUOTE=QUOTE,即kAB=kBC.故三點(diǎn)共線,所以①正確.②中由a3+a7=-6,a1=-11得,-22+8d=-6,∴d=2>0,故{an}是遞增的等差數(shù)列,故Sn無最大值,②錯(cuò).③中,當(dāng){an}是搖擺數(shù)列時(shí),如{an}為1,-1,1,-1,…時(shí),m取偶數(shù)時(shí)Sm=0,故③錯(cuò).④中,n≥2時(shí),由已知Sn=a1+qSn-1,∴Sn+1-Sn=q(Sn-Sn-1),即an+1=qan,∴QUOTE=q.當(dāng)n=1時(shí),a2+a1=a1+qa1,故QUOTE=q,故{an}為等比數(shù)列.答案:①④12.【解析】設(shè)定值為M,則an+an+1+an+2=M,進(jìn)而an+1+an+2+an+3=M,后式減去前式得an+3=an,即數(shù)列{an}是以3為周期的數(shù)列.由a7=2,可知a1=a4=a7=…=a100=2,共34項(xiàng),其和為68;由a9=3,可得a3=a6=…=a99=3,共33項(xiàng),其和為99;由a98=4,可得a2=a5=…=a98=4,共33項(xiàng),其和為132.故數(shù)列{an}的前100項(xiàng)的和S100=68+99+132=299.答案:29913.【解析】(1)設(shè)等差數(shù)列{log2(an-1)}的公差為d.由a1=3,a3=9得2(log22+d)=log22+log28,即d=1.所以log2(an-1)=1+(n-1)×1=n,即an=2n+1.(2)由于QUOTE=QUOTE=QUOTE,所以Sn=QUOTE+QUOTE+…+QUOTE=QUOTE+QUOTE+QUOTE+…+QUOTE=QUOTE=1-QUOTE.14.【解析】(1)2a1+3a2=2a1+3(a1+d)=5a1+3d=11,2a3=a2+a6-4,即2(a1+2d)=a1+d+a1+5d-4,得d=2,則a1=1,故an=2n-1.(2)由(1)得Sn=n2,∴bn=QUOTE=QUOTE=QUOTE=QUOTE=QUOTE(QUOTE-QUOTE),Tn=QUOTE(QUOTE-QUOTE+QUOTE-QUOTE+QUOTE-QUOTE+…+QUOTE-QUOTE+QUOTE-QUOTE)=QUOTE(QUOTE+QUOTE-QUOTE-QUOTE)<QUOTE(n∈N*).15.【解析】(1)在等差數(shù)列{an}中,a3+a5=10=2a4,∴a4=5.S5=15=5a3,∴a3=3,則等差數(shù)列{an}的公差為2.∴an=2n-3.(2)bn=(QUOTE)n·an=QUOTE,Tn=QUOTE+QUOTE+QUOTE+…+QUOTE,QUOTETn=QUOTE+QUOTE+QUOTE+…+QUOTE+QUOTE,QUOTETn=QUOTE+2·(QUOTE+QUOTE+…+QUOTE)-QUOTETn=-1+(1+QUOTE+QUOTE+QUOTE+…+QUOTE)-QUOTE=1-QUOTE.16.【解析】QUOTE=QUOTE=QUOTE=QUOTE=QUOTE-QUOTE,k=1,2,3,…,n故QUOTE+QUOTE+…+QUOTE=(QUOTE-QUOTE)+(QUOTE-QUOTE)+…+[QUOTE-QUO