高考數(shù)學(xué)二輪總復(fù)習(xí)題型專項(xiàng)集訓(xùn)題型練4大題專項(xiàng)（二）文

題型練4大題專項(xiàng)(二)數(shù)列的通項(xiàng)、求和問題1.(2022新高考Ⅱ,17)已知{an}為等差數(shù)列,{bn}為公比為2的等比數(shù)列,且a2b2=a3b3=b4a4.(1)證明:a1=b1;(2)求集合{k|bk=am+a1,1≤m≤500}中元素的個(gè)數(shù).2.設(shè)數(shù)列{an}的前n項(xiàng)和為Sn,已知a1=1,Sn=n+12a(1)求數(shù)列{an}的通項(xiàng)公式;(2)設(shè)bn=2an,數(shù)列{bn}的前n項(xiàng)和為Tn,求證:Tn·Tn+2<3.已知公差不為零的等差數(shù)列{an}滿足a3=4,且a2,a1,a3成等比數(shù)列.(1)求數(shù)列{an}的通項(xiàng)公式;(2)若數(shù)列{an3n1}的前n項(xiàng)和為Sn,求使Sn≤20成立的最小正整數(shù)n.4.(2022山東煙臺(tái)模擬)已知{an}是公差為2的等差數(shù)列,a1>0,且a4是2a2和a52的等比中項(xiàng).(1)求{an}的通項(xiàng)公式;(2)設(shè)數(shù)列{bn}滿足b1a1+b2a2+…+bnan=25.已知函數(shù)f(x)=7x+5x+1,數(shù)列{an}滿足:2an+12an+an+1an=0,且anan+1≠0.在數(shù)列{bn}中,b1=f(0),且bn=f((1)求證:數(shù)列1a(2)求數(shù)列{|bn|}的前n項(xiàng)和Tn.6.已知等比數(shù)列{an}的公比q>1,且a3+a4+a5=28,a4+2是a3,a5的等差中項(xiàng).數(shù)列{bn}滿足b1=1,數(shù)列{(bn+1bn)an}的前n項(xiàng)和為2n2+n.(1)求q的值;(2)求數(shù)列{bn}的通項(xiàng)公式.答案：1.(1)證明:設(shè)等差數(shù)列{an}的公差為d,由a2b2=a3b3,等比數(shù)列{bn}的公比為2,得a1+d2b1=a1+2d4b1,得d=2b1.由a2b2=b4a4,等比數(shù)列{bn}的公比為2,得a1+d2b1=8b1(a1+3d),所以a1+2b12b1=8b1(a1+6b1),所以a1=b1.(2)解:由bk=am+a1,可得b1·2k1=a1+(m1)d+a1,由(1)知d=2b1=2a1,所以b1·2k1=b1+(m1)·2b1+b1,所以2k1=2m.又1≤m≤500,∴2≤2k1≤1000.∴2≤k≤10.又k∈Z,故集合{k|bk=am+a1,1≤m≤500}中元素的個(gè)數(shù)為9.2.(1)解:∵Sn=n+12an,∴2Sn=(n+1)a∴2Sn+1=(n+2)an+1,∴2Sn+12Sn=(n+2)an+1(n+1)an,即2an+1=(n+2)an+1(n+1)an,∴(n+1)an=nan+1,即an∴ann是以a11=1為首項(xiàng)的常數(shù)列,∴(2)證明:由(1)知bn=2n,∴Tn=2(1-2n∴T

n+12Tn·Tn+2=4(2n+11)24(2n1)·(2n+21)=4(22n+22n+2+122n+2+2n+2n+21)=4×2n=2n+2>0,∴Tn·Tn+23.解:(1)設(shè)等差數(shù)列{an}的公差為d(d≠0).因?yàn)閍2,a1,a3成等比數(shù)列,所以a12=a2a3,即(a32d)2=(a3d)a3,所以(42d)2=4(4d又d≠0,所以d=3.所以a1=a32d=2.所以an=23(n1)=3n+5.(2)由(1)得an3n1=(3n+5)3n1,所以Sn=(230)+(131)+(432)+…+[(3n+5)3n1]=[2+(1)+(4)+…+(3n+5)](30+31+32+…+3n1)=n[所以S1=1,S2=3,S1>S2.由f(x)=7x-3x22在區(qū)間[2,+∞)上單調(diào)遞減,g(x)=3x-12在區(qū)間[2,+又S2<S1,所以{Sn}是遞減數(shù)列.又S3=16,S4=50,所以使Sn≤20成立的最小正整數(shù)n為4.4.解:(1)依題意,{an}是公差為2的等差數(shù)列,a1>0,且a4是2a2和a52的等比中項(xiàng),所以a42=2a2·(a52),所以(a1+6)2=2(a1+2)·(a1+6),解得a1=2或a1所以an=2+2(n1)=2n.(2)依題意,b1a1+b2a2+…當(dāng)n=1時(shí),b1a1=22,b當(dāng)n≥2時(shí),b1a1+b2a2+①②得bnan=2n,所以bn=2n·2n=n·2所以bn=8所以Tn=8+2×23+3×24+…+n·2n+1,③2Tn=16+2×24+3×25+…+n·2n+2,④③④得Tn=23+24+…+2n+1n·2n+2=8(1-2n-1)1-2n·2所以Tn=(n1)·2n+2+8.5.(1)證明:∵2an+12an+an+1an=0,∴1a故數(shù)列1an是以1(2)解:∵b1=f(0)=5,∴7(a1-1)+5a1-1+1=5,7a12=5a1,∴a1=∴an=2n+1,bn=7an-2an當(dāng)n≤6時(shí),Tn=n2(5+6n)=n當(dāng)n≥7時(shí),Tn=15+n-62(1+n6)故Tn=n6.解:(1)由a4+2是a3,a5的等差中項(xiàng),得a3+a5=2a4+4,所以a3+a4+a5=3a4+4=28,解得a4=8.由a3+a5=20,得8q+1解得q=2或q=12,因?yàn)閝>1,所以q=2(2)設(shè)cn=(bn+1bn)an,數(shù)列{cn}前n項(xiàng)和為Sn,由cn=S1,n=1,Sn-由(1)可知an=2n1,所以bn+1bn=(4n1)·12故bnbn1=(4n5)·12n-bnb1=(bnbn1)+(bn1bn2)+…+(b3b2)+(b2b1)=(4n5)·12n-2+(4n9)·12n-3+設(shè)Tn=3