課時規(guī)范練29等比數(shù)列及其前n項和基礎(chǔ)穩(wěn)固組1.已知等比數(shù)列{a}知足a=,aa=4(a-1),則a=( )n13542A.2B.1C.D.2.在正項等比數(shù)列{an}中,a2,a48是方程2x2-7x+6=0的兩個根,則a1·a2·a25·a48·a49的值為()A.B.9C.±9D.353.(2017安徽黃山市二模)已知數(shù)列{an}的前n項和為Sn,且a1=2,an+1=Sn+1(n∈N*),則S5=()A.31B.42C.37D.474.設(shè)首項為1,公比為的等比數(shù)列{a}的前n項和為S,則()nnA.Sn=2an-1B.Sn=3an-2C.Sn=4-3anD.Sn=3-2an5.(2017全國Ⅲ)等差數(shù)列{an}的首項為1,公差不為0.若a2,a3,a6成等比數(shù)列,則{an}前6項的和為()A.-24B.-3C.3D.86.設(shè)等比數(shù)列{an}的前n項和為Sn.若S2=3,S4=15,則S6=()A.31B.32C.63D.647.設(shè)數(shù)列{an}是首項為a1,公差為-1的等差數(shù)列,Sn為其前n項和.若S1,S2,S4成等比數(shù)列,則a1的值為.8.(2017北京)若等差數(shù)列{a}和等比數(shù)列知足111,448,則=.nn9.(2017江蘇,9)等比數(shù)列{a}的各項均為實數(shù),其前n項和為S.已知S=,S=,則nn36a8=.10.(2017n1nnn+1n*河南新鄉(xiāng)二模,文17)在數(shù)列{a}中,a=,{a}的前n項和S知足S-S=(n∈N).(1)求數(shù)列{an}的通項公式an以及前n項和Sn;(2)若12,13,(23)成等差數(shù)列,務(wù)實數(shù)的值S+SS+SmS+Sm.?導(dǎo)學(xué)號24190754?綜合提高組11.(2017四川廣元二診)已知數(shù)列{a}的前n項和為S,且對隨意正整數(shù)n都有a=S+2建立.若nnnnbn=log2an,則b1008=()A.2017B.2016C.2015D.201412.設(shè)等比數(shù)列{an}知足a1+a3=10,a2+a4=5,則a1·a2·a3··an的最大值為.13.已知{a}是公差為3的等差數(shù)列,數(shù)列知足b1=1,b2=,ab1+b1=nb.nnnn+n+n求{an}的通項公式;求{bn}的前n項和.創(chuàng)新應(yīng)用組14.已知數(shù)列{an}的前n項和為Sn,在數(shù)列{bn}中,b1=a1,bn=an-an-1(n≥),且an+Sn=n.設(shè)cn=an-1,求證:{cn}是等比數(shù)列;求數(shù)列{bn}的通項公式.答案：1.C354(41),4(41),∵aa=a-∴=a-解得a4=2.又a4=a1q3,且a1=,∴q=2,a2=a1q=.2.B∵a2,a48是方程2x2-7x+6=0的兩個根,∴a2·a48=3.又a1·a49=a2·a48==3,a25>0,a1·a2·a25·a48·a49==9.3.D∵an+1=Sn+1(n∈N*),Sn+1-Sn=Sn+1(n∈N*),Sn+1+1=2(Sn+1)(n∈N*),∴數(shù)列{Sn+1}是首項為3,公比為2的等比數(shù)列.則S5+1=3×24,解得S5=47.4.n-)--n---5.A設(shè)等差數(shù)列的公差為,則d≠0,26,即(12)2(1)(15),解得d=-2,所以d=a·a+d=+d+dS6=6×1+×(-2)=-24,應(yīng)選A.6.C∵S2=3,S4=15,∴由等比數(shù)列前n項和的性質(zhì),得S2,S4-S2,S6-S4成等比數(shù)列,(S4-S2)2=S2(S6-S4),即(15-3)2=3(S6-15),解得S6=63,應(yīng)選C.7.-由已知得S1=a1,S2=a1+a2=2a1-1,S4=4a1+×(-1)=4a1-6,而S1,S2,S4成等比數(shù)列,(2a1-1)2=a1(4a1-6),整理,得2a1+1=0,解得a1=-.8.1設(shè)等差數(shù)列{an}的公差為d,等比數(shù)列{bn}的公比為q,由題意知-1+3d=-q3=8,-,,即解得-,--故=1.-)9.32設(shè)該等比數(shù)列的公比為q,則S6-S3==14,即a4+a5+a6=14.①∵S3=,∴a1+a2+a3=.由①得(a1+a2+a3)q3=14,q3==8,即q=2.a1+2a1+4a1=,a1=,a8=a1·q7=×27=32.10.解(1)n+1n+1n=,∵a=S-S∴當(dāng)n≥2時,an=.又a1=,∴當(dāng)n=1時上式也建立.∴a=,n∴Sn=-=1-.-由(1)可得:S1=,S2=,S3=.∵S1+S2,S1+S3,m(S2+S3)成等差數(shù)列,∴+m=2,解得m=.11.A在an=Sn+2中,令n=1得a1=8,∵an=Sn+2建立,an+1=Sn+1+2建立,兩式相減得an+1-an=an+1,an+1=4an,又a1≠0,∴數(shù)列{an}為等比數(shù)列,an=8·4n-1=22n+1,∴bn=log2an=2n+1,∴b1008=2017,應(yīng)選A.12.64由已知a1+a3=10,a2+a4=(a1+a3)q=5,得q=,所以a1=8,n?-)-,n所以當(dāng)n=3或n=4時,a1·a2·a3··an取最大值為-6=64.=213.解(1)由已知,得a1b2+b2=b1,由于b=1,b=,所以a=2.121所以數(shù)列{a}是首項為2,公差為3的等差數(shù)列,通項公式為31nn(2)由(1)和ab+b=nb,得bn+1=,所以是首項為1,公比為的等比數(shù)列.nn+1n+1nn記的前n項和為S,則S=--.-nnn14.(1)證明nn,①∵a+S=n∴a+S=n+.②n+1n+11-①得an+1-an+an+1=1,2an+1=an+1,2(an+1-1)=an-1,∴

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,-{an-1}是等比數(shù)列.又a1+a1=1,∴a1=,∵首項c1=a1-