2024屆新教材二輪復(fù)習(xí) 等比數(shù)列的前n項(xiàng)和 作業(yè)

等比數(shù)列的前n項(xiàng)和A級(jí)必備知識(shí)基礎(chǔ)練1.設(shè)Sn為等比數(shù)列{an}的前n項(xiàng)和,若公比q=-3,則S4S2=A.10 B.9C.-8 D.-52.等比數(shù)列{an}的前n項(xiàng)和為Sn,若S3=15,a3=5,則公比q的值為()A.-12 B.C.-12或1 D.13.已知在等比數(shù)列{an}中,an=3·2n-1,則a1+a3+a5+…+a2k-1=()A.4k-1 B.3(2k-1)C.2(4k-1) D.3(4k-1)4.已知等比數(shù)列{an}的前n項(xiàng)和為Sn,且滿足a1=2,S4-S3=1627,則數(shù)列{an}的前4項(xiàng)和為(A.13027 B.C.6527 D.5.若正項(xiàng)等比數(shù)列{an}滿足a1a3=116,2a4+a3=a2,則Sn=1a1-1a2+…+(-1)A.23[1+(-2)n] B.23(1-2C.23(1+2n) D.23[1-(-2)6.我國(guó)古代數(shù)學(xué)名著《算法統(tǒng)宗》中有如下問(wèn)題:“遠(yuǎn)望巍巍塔七層,紅光點(diǎn)點(diǎn)倍加增,共燈三百八十一,請(qǐng)問(wèn)尖頭幾盞燈?”根據(jù)詩(shī)詞的意思,可得塔的最底層共有燈()A.192盞 B.128盞 C.3盞 D.1盞7.等比數(shù)列{an}的前n項(xiàng)和為Sn,已知S3=a2+10a1,a5=9,則a1=.

8.記Sn為等比數(shù)列{an}的前n項(xiàng)和,且Sn≠0,已知a1=1,S4=5S2.(1)求{an}的通項(xiàng)公式;(2)若Sm=43,求m.B級(jí)關(guān)鍵能力提升練9.已知各項(xiàng)均為正數(shù)的等比數(shù)列{an}的前4項(xiàng)和為15,且a5=3a3+4a1,則a3=()A.16 B.8 C.4 D.210.已知數(shù)列{an}為等比數(shù)列,a1=1,q=2,且第m項(xiàng)至第n(m<n)項(xiàng)的和為112,則m+n的值為()A.11 B.12 C.13 D.1411.(2023新高考Ⅱ,8)記Sn為等比數(shù)列{an}的前n項(xiàng)和,若S4=-5,S6=21S2,則S8=()A.120 B.85 C.-85 D.-12012.已知數(shù)列{an}是首項(xiàng)a1=14的等比數(shù)列,其前n項(xiàng)和為Sn,S3=316,若am=-1512,則m的值為13.已知等比數(shù)列{an}的前n項(xiàng)和為Sn,S4=2S2,則數(shù)列{an}的公比q=.

14.設(shè){an}為等比數(shù)列,其前n項(xiàng)和為Sn,a2=2,S2-3a1=0,則{an}的通項(xiàng)公式是;Sn+an>48,則n的最小值為.

15.已知數(shù)列{an}滿足a1=1,nan+1=3(n+1)an.(1)設(shè)bn=ann,求證:數(shù)列{b(2)求數(shù)列{an}的前n項(xiàng)和Sn.C級(jí)學(xué)科素養(yǎng)創(chuàng)新練16.條件①:設(shè)數(shù)列{an}的前n項(xiàng)和為Sn,且Sn=2n+k(n∈N+,k∈R),a1=1.條件②:對(duì)?n∈N+,有an+1an=q>1(q為常數(shù)),a3=4,并且a2-1,a3,a4-在數(shù)列{an}中,.

(1)求數(shù)列{an}的通項(xiàng)公式an;(2)記Tn=a1+2a2+3a3+…+nan,求T10的值.第1課時(shí)等比數(shù)列的前n項(xiàng)和1.A由q=-3,可知S4S22.C∵S3=a1+a2+a3=15,a3=5,∴a1+a2=10.∴a1(1+q)=10,而a1q2=5,即1+q=2q2,解得q=-12或1.故選C3.A∵{an}是首項(xiàng)為3,公比為2的等比數(shù)列,∴{a2k-1}是首項(xiàng)為3,公比為4的等比數(shù)列,∴a1+a3+…+a2k-1=3(1-4k)1-4.A根據(jù)題意,設(shè)等比數(shù)列{an}的公比為q,因?yàn)閍1=2,S4-S3=1627,所以a4=1627.則q3=a4a故數(shù)列{an}的前4項(xiàng)和S4=a1(5.D由題意,a1a3=a22=116,得a令{an}的公比為q,且q>0,由2a4+a3=a2,得2q2+q-1=0,解得q=12(負(fù)值舍去).∴a1=12,∴an=令bn=(-1)n+11an,則bn=-(-2)∴Sn=b1+b2+…+bn=2[1-(-2)n]16.A設(shè)這個(gè)塔頂層有x盞燈,則七層塔從頂?shù)降酌繉訜舻谋K數(shù)構(gòu)成一個(gè)首項(xiàng)為x,公比為2的等比數(shù)列,且其前7項(xiàng)和為381,所以x(27-1)2-1=381,解得x=7.19S3=a1+a2+a3=a2+10a1,即a1q2=9a1,即q2=9又9=a5=a1q4,∴a1=198.解(1)設(shè){an}的公比為q,由S4=5S2得a1+a2+a3+a4=5(a1+a2),整理得a3+a4=4(a1+a2),因?yàn)閍1+a2≠0,所以q2=4,所以q=2或q=-2,故an=2n-1或an=(-2)n(2)若an=2n-1,則Sn=2n-1,由Sm=43,得2m=44,此方程沒(méi)有正整數(shù)解.若an=(-2)n-1,則Sn=1-(-2)n3,由Sm=43,得(-2)m綜上,m=7.9.C設(shè)等比數(shù)列{an}的公比為q(q>0),則由前4項(xiàng)和為15,且a5=3a3+4a1,得a1+a1q+a1q2+10.B由已知得1×(即2m-1·(2n-m+1-1)=24×7.又因?yàn)?n-m+1-1為奇數(shù),所以2m-1=2411.C如果q=1,則S6=6a1,S2=2a1,不滿足S6=21S2,所以q≠1.由S6=21S2得a1(1-所以1-q6=(1-q2)(1+q2+q4)=21(1-q2),則q4+q2-20=0,解得q2=4或q2=-5(舍去).由已知S4=a1(1-q4)1-q=-12.8設(shè)數(shù)列{an}的公比為q,因?yàn)镾3=316,所以a1+a2+a3=316,即a1(1+q+q2)=因?yàn)閍1=14,所以1+4q+4q2=0,解得q=-1所以am=14×(-12)m-1=-1512,解得13.±1由S4=2S2,得(a1+a2+a3+a4)=2(a1+a2),即a3+a4=a1+a2,進(jìn)而可得q2=1,解得q=±1.14.an=2n-16設(shè)等比數(shù)列{an}的公比為q,則a2=a1q=2,S2-3a1=a2+a1-3a1=0,解得a1=1,q=2,故an=1×2n-1=2n-1,Sn=1·(1-2nSn+an=2n-1+2n-1>48,即3·2n-1>49,故n的最小值為6.15.(1)證明由nan+1=3(n+1)an可得an因?yàn)閎n=ann,a1=1,所以bn+1=3bn,b1=故{bn}是首項(xiàng)為1,公比為3的等比數(shù)列.(2)解由(1)可得ann=bn=3n-1,則an=n·3n-Sn=1·30+2·31+3·32+…+n·3n-1,3Sn=1·31+2·32+3·33+…+(n-1)·3n-1+n·3n,上面兩式相減可得-2Sn=30+31+32+…+3n-1-n·3n=1-3n1-則Sn=(216.解選條件①.(1)由S1=2+k=a1=1得k=-1,∴Sn=2n-1.當(dāng)n≥2時(shí),an=Sn-Sn-1=2n-1,又a1=1,符合上式,∴數(shù)列{an}的通項(xiàng)公式為an=2n-1.(2)∵T10=1+2×2+3×22+4×23+…+10×29,∴2T10=2+2×22+3×23+4×24+…+9×29+10×210,兩式相減,得-T10=1+2+22+23+…+29-10×210=210-12-1∴T10=9×210+1.選條件②.(1)由an+1an=q知數(shù)列{a且a2=a3q=4q,a4=a由已知可得2a3=a2-1+a4-1,