【全程復(fù)習(xí)方略】2020年人教A版數(shù)學(xué)理(廣東用)課時(shí)作業(yè)：第五章-第四節(jié)數(shù)-列-求-和

溫馨提示：此套題為Word版，請(qǐng)按住Ctrl,滑動(dòng)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)(三十三)一、選擇題1.已知數(shù)列{an},若點(diǎn)(n,an)(n∈N*)在經(jīng)過點(diǎn)(5,3)的定直線l上,則數(shù)列{an}的前9項(xiàng)和S9=()(A)9 (B)10 (C)18 (D)272.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.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.(2021·東莞模擬)已知數(shù)列{an}的通項(xiàng)公式是an=(-1)n(n+1),則a1+a2+a3+…+a10=()(A)-55 (B)-5 (C)5 (D)555.(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.(2021·佛山模擬)在數(shù)列{an}中,a1=1,a2=2,且an+2-an=1+(-1)n(n∈N*),則S100=.11.(2021·湛江模擬)在等差數(shù)列{an}中,首項(xiàng)a1=0,公差d≠0,若ak=a1+a2+a3+…+a7,則k=.12.(2021·哈爾濱模擬)在數(shù)列{an}中,若對(duì)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.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*).14.(2021·湖州模擬)設(shè){an}是等差數(shù)列,{bn}是各項(xiàng)都為正數(shù)的等比數(shù)列,且a1=b1=1,a3+b5=21,a5+b3=13.(1)求{an},{bn}的通項(xiàng)公式.(2)求數(shù)列{QUOTE}的前n項(xiàng)和Sn.15.(力氣挑戰(zhàn)題)已知數(shù)列{an}的首項(xiàng)為a1=1,其前n項(xiàng)和為Sn,且對(duì)任意正整數(shù)n有n,an,Sn成等差數(shù)列.(1)求證:數(shù)列{Sn+n+2}成等比數(shù)列.(2)求數(shù)列{an}的通項(xiàng)公式.答案解析1.【思路點(diǎn)撥】(n,an)在直線上說明數(shù)列{an}為等差數(shù)列.【解析】選D.點(diǎn)(n,an)(n∈N*)在經(jīng)過點(diǎn)(5,3)的定直線l上,a5=3,依據(jù)等差數(shù)列性質(zhì)得:S9=9a5=27.2.【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=(-1)n(n+1),得a1+a2+a3+…+a10=-2+3-4+5-6+…-10+11=5×1=5,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)成某等比數(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(duì)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】∵a2n+1-a2n-1=0,a2n+2-a2n=2,∴數(shù)列{an}的奇數(shù)項(xiàng)為常數(shù)列,偶數(shù)項(xiàng)是公差為2的等差數(shù)列,∴S100=(a1+a3+…+a99)+(a2+a4+…+a100)=(1+1+…+1)+(2+4+…+100)=50+50×51=2600.答案:260011.【解析】ak=a1+a2+a3+…+a7=7a1+QUOTEd=21d=a22,∴k=22.答案:2212.【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)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*).【方法技巧】裂項(xiàng)相消法的應(yīng)用技巧裂項(xiàng)相消法的基本思想是把數(shù)列的通項(xiàng)an分拆成an=bn+1-bn或者an=bn-bn+1或者an=bn+2-bn等,從而達(dá)到在求和時(shí)逐項(xiàng)相消的目的,在解題中要擅長依據(jù)這個(gè)基本思想變換數(shù)列an的通項(xiàng)公式,使之符合裂項(xiàng)相消的條件.在裂項(xiàng)時(shí)確定要留意把數(shù)列的通項(xiàng)分拆成的兩項(xiàng)確定是某個(gè)數(shù)列中的相鄰的兩項(xiàng)或者是等距離間隔的兩項(xiàng),只有這樣才能實(shí)現(xiàn)逐項(xiàng)相消后剩下幾項(xiàng),達(dá)到求和的目的.14.【解析】(1)設(shè){an}的公差為d,{bn}的公比為q,則依題意有q>0且QUOTE解得QUOTE所以an=1+(n-1)d=2n-1,bn=qn-1=2n-1.(2)QUOTE=QUOTE,Sn=1+QUOTE+QUOTE+…+QUOTE+QUOTE,①2Sn=2+3+QUOTE+…+QUOTE+QUOTE.②②-①,得Sn=2+2+QUOTE+QUOTE+…+QUOTE-QUOTE=2+2×(1+QUOTE+QUOTE+…+QUOTE)-QUOTE,=2+2×QUOTE-QUOTE=6-QUOTE.【變式備選】(2022·石家莊模擬)已知各項(xiàng)都不相等的等差數(shù)列{an}的前6項(xiàng)和為60,且a6為a1和a21的等比中項(xiàng).(1)求數(shù)列{an}的通項(xiàng)公式.(2)若數(shù)列{bn}滿足bn+1-bn=an(n∈N*),且b1=3,求數(shù)列{QUOTE}的前n項(xiàng)和Tn.【解析】(1)設(shè)等差數(shù)列{an}的公差為d(d≠0),則QUOTE解得QUOTE∴an=2n+3.(2)由bn+1-bn=an,∴bn-bn-1=an-1(n≥2,n∈N*),bn=(bn-bn-1)+(bn-1-bn-2)+…+(b2-b1)+b1=an-1+an-2+…+a1+b1=n(n+2),當(dāng)n=1時(shí),b1=3也適合上式,∴bn=n(n+2)(n∈N*).∴QUOTE=QUOTE=QUOTE(QUOTE-QUOTE),Tn=QUOTE(1-QUOTE+QUOTE-QUOTE+…+QUOTE-QUOTE)=QUOTE(QUOTE-QUOTE-QUOTE)=QUOTE.15.【解析】(1)由于n,an,Sn成等差數(shù)列,所以2an=Sn+n,由當(dāng)n≥2時(shí),an=S