高考第一輪文科數(shù)學(xué)（人教A版）解答題專項(xiàng)三　數(shù)列-試卷下載-教習(xí)網(wǎng)

(1)若S4-2a2a3+6=0,求Sn;
(2)若對于每個n∈N*,存在實(shí)數(shù)cn,使an+cn,an+1+4cn,an+2+15cn成等比數(shù)列,求d的取值范圍.
2.已知數(shù)列{an}的前n項(xiàng)和為Sn,數(shù)列Snn是首項(xiàng)為12,公差為14的等差數(shù)列,若[x]表示不超過x的最大整數(shù),如[0.5]=0,[lg 499]=2.
(1)求數(shù)列{an}的通項(xiàng)公式;
(2)若bn=[lg an],求數(shù)列{bn}的前2 021項(xiàng)的和.
3.(2022河南鄭州一模)已知等差數(shù)列{an}的公差為d(d≠0),前n項(xiàng)和為Sn,現(xiàn)給出下列三個條件:①S1,S2,S4成等比數(shù)列;②S4=16;③S8=4(a8+1).請你從這三個條件中任選兩個解答下列問題.
(1)求{an}的通項(xiàng)公式;
(2)若bn-bn-1=4an(n≥2),且b1=3,求數(shù)列1bn的前n項(xiàng)和Tn.
4.已知數(shù)列{an}的前n項(xiàng)和為Sn,a1=1,且數(shù)列Snn是以1為公差的等差數(shù)列.
(1)求數(shù)列1anan+1的前n項(xiàng)和Tn;
(2)設(shè)等比數(shù)列{cn}的首項(xiàng)為2,公比為q(q>0),其前n項(xiàng)和為Pn,若存在正整數(shù)m,使得S3是Sm與P3的等比中項(xiàng),求q的值.
5.Sn為等比數(shù)列{an}的前n項(xiàng)和,已知a4=9a2,S3=13,且公比q>0.
(1)求an及Sn.
(2)是否存在常數(shù)λ,使得數(shù)列{Sn+λ}是等比數(shù)列?若存在,求出λ的值;若不存在,請說明理由.
6.已知等比數(shù)列{an}的公比為λ(λ>1),a1=1,數(shù)列{bn}滿足bn+1-bn=an+1-λ,b1=1λ-1.
(1)求數(shù)列{bn}的通項(xiàng)公式;
(2)規(guī)定:[x]表示不超過x的最大整數(shù),如[-1.2]=-2,[2.1]=2.若λ=2,cn=1bn+2n-2,記Tn=c1+c2+c3+…+cn(n≥2),求Tn2-2Tn+2Tn-1的值,并指出相應(yīng)n的取值范圍.
參考答案
解答題專項(xiàng)三數(shù)列
1.解(1)數(shù)列{an}是首項(xiàng)a1=-1的等差數(shù)列且d>1.
∵S4-2a2a3+6=0,
∴4a1+4×32d-2(a1+d)(a1+2d)+6=0.
把a(bǔ)=-1代入得-4d2+12d=0,
解得d=3或d=0(舍去),
∴Sn=na1+n(n-1)2d=3n2-5n2.
(2)∵對每個n∈N*,存在實(shí)數(shù)cn使得an+cn,an+1+4cn,an+2+15cn成等比數(shù)列,
∴(an+1+4cn)2=(an+cn)(an+2+15cn),
an+12+8an+1cn+16cn2=anan+2+an+2cn+15ancn+15cn2,
cn2+(8an+1-an+2-15an)cn+an+12-anan+2=0,
而8an+1-an+2-15an=8(a1+nd)-[a1+(n+1)d]-15[a1+(n-1)d]=8a1+8nd-a1-(n+1)d-15a1-15(n-1)d=-8a1+(8n-n-1-15n+15)d=8+(14-8n)d,
an+12-an·an+2=(an+d)2-an(an+2d)=d2,
∴cn2+[8+(14-8n)d]cn+d2=0,
對此式,Δ=[8+(14-8n)d]2-4d2≥0,
[8+(14-8n)d+2d][8+(14-8n)d-2d]=[(16-8n)d+8][(12-8n)d+8]≥0,
[(2-n)d+1][(3-2n)d+2]≥0,
n=1時,顯然成立;
n=2時,-d+2≥0,d≤2;
n≥3時,原式=[(n-2)d-1][(2n-3)d-2]>0恒成立.
∴1