高考數(shù)學(xué)理二輪專題復(fù)習(xí)突破精練：專題對點(diǎn)練14　數(shù)列與數(shù)列不等式的證明及數(shù)列中的存在性問題 Word版含解析

專題對點(diǎn)練14數(shù)列與數(shù)列不等式的證明及數(shù)列中的存在性問題專題對點(diǎn)練第19頁 1.若數(shù)列an滿足:a1=23,a2=2,3(an+1-2an+an-1)=2.(1)證明:數(shù)列an+1-an是等差數(shù)列;(2)求使1a1+1a2+1a3+1an>52成立的最小的正整數(shù)n.(1)證明 由3(an+1-2an+an-1)=2可得an+1-2an+an-1=23,即(an+1-an)-(an-an-1)=23,故數(shù)列an+1-an是以a2-a1=43為首項(xiàng),23為公差的等差數(shù)列.(2)解 由(1)知an+1-an=43+23(n-1)=23(n+1),于是累加求和得an=a1+23(2+3+n)=13n(n+1),1an=31n-1n+1.1a1+1a2+1a3+1an=3-3n+1>52,n>5.最小的正整數(shù)n為6.2.(2017廣東揭陽二模,理17)已知數(shù)列an,a1=1,an+1=2(n+1)ann+n+1.(1)求證:數(shù)列ann+1是等比教列;(2)求數(shù)列an的前n項(xiàng)和Sn.(1)證明 an+1=2(n+1)ann+n+1,an+1n+1=2×ann+1,an+1n+1+1=2×ann+1,數(shù)列ann+1是等比教列,公比為2,首項(xiàng)為2.(2)解 由(1)可得ann+1=2n,可得an=n·2n-n.設(shè)數(shù)列n·2n的前n項(xiàng)和為Tn,則Tn=2+2×22+3×23+n·2n,2Tn=22+2×23+(n-1)·2n+n·2n+1,兩式相減可得-Tn=2+22+2n-n·2n+1=2(1-2n)1-2-n·2n+1,Tn=(n-1)·2n+1+2.Sn=(n-1)·2n+1+2-n(n+1)2.3.已知數(shù)列an的前n項(xiàng)和Sn=1+an,其中0.(1)證明an是等比數(shù)列,并求其通項(xiàng)公式;(2)若S5=3132,求.解 (1)由題意得a1=S1=1+a1,故1,a1=11-,a10.由Sn=1+an,Sn+1=1+an+1得an+1=an+1-an,即an+1(-1)=an.由a10,0得an0,所以an+1an=-1.因此an是首項(xiàng)為11-,公比為-1的等比數(shù)列,于是an=11-1n-1.(2)由(1)得Sn=1-1n.由S5=3132得1-15=3132,即-15=132.解得=-1.4.(2017吉林白山二模,理17)在數(shù)列an中,設(shè)f(n)=an,且f(n)滿足f(n+1)-2f(n)=2n(nN*),且a1=1.(1)設(shè)bn=an2n-1,證明數(shù)列bn為等差數(shù)列;(2)求數(shù)列an的前n項(xiàng)和Sn.(1)證明 由已知得an+1=2an+2n,bn+1=an+12n=2an+2n2n=an2n-1+1=bn+1,bn+1-bn=1.又a1=1,b1=1,bn是首項(xiàng)為1,公差為1的等差數(shù)列.(2)解 由(1)知,bn=an2n-1=n,an=n·2n-1.Sn=1+2×21+3×22+n·2n-1,2Sn=1×21+2×22+(n-1)·2n-1+n·2n,兩式相減得-Sn=1+21+22+2n-1-n·2n=2n-1-n·2n=(1-n)2n-1,Sn=(n-1)·2n+1.5.設(shè)數(shù)列an的前n項(xiàng)和為Sn,且(3-m)Sn+2man=m+3(nN*),其中m為常數(shù),且m-3.(1)求證:an是等比數(shù)列;(2)若數(shù)列an的公比q=f(m),數(shù)列bn滿足b1=a1,bn=32f(bn-1)(nN*,n2),求證:1bn為等差數(shù)列,并求bn.證明 (1)由(3-m)Sn+2man=m+3,得(3-m)Sn+1+2man+1=m+3,兩式相減,得(3+m)an+1=2man.m-3,an+1an=2mm+3,an是等比數(shù)列.(2)由(3-m)Sn+2man=m+3,得(3-m)S1+2ma1=m+3,即a1=1,b1=1.數(shù)列an的公比q=f(m)=2mm+3,當(dāng)n2時(shí),bn=32f(bn-1)=32·2bn-1bn-1+3,bnbn-1+3bn=3bn-1,1bn-1bn-1=13.1bn是以1為首項(xiàng),13為公差的等差數(shù)列,1bn=1+n-13=n+23.又1b1=1也符合,bn=3n+2.6.已知數(shù)列an的前n項(xiàng)和為Sn,a1=-2,且滿足Sn=12an+1+n+1(nN*).(1)求數(shù)列an的通項(xiàng)公式;(2)若bn=log3(-an+1),求數(shù)列1bnbn+2的前n項(xiàng)和Tn,并求證Tn<34.(1)解 Sn=12an+1+n+1(nN*),當(dāng)n=1時(shí),-2=12a2+2,解得a2=-8.當(dāng)n2時(shí),an=Sn-Sn-1=12an+1+n+1-12an+n,即an+1=3an-2,可得an+1-1=3(an-1).當(dāng)n=1時(shí),a2-1=3(a1-1)=-9,數(shù)列an-1是等比數(shù)列,首項(xiàng)為-3,公比為3.an-1=-3n,即an=1-3n.(2)證明 bn=log3(-an+1)=n,1bnbn+2=121n-1n+2.Tn=121-13+12-14+13-15+1n-1-1n+1+1n-1n+2=121+12-1n+1-1n+2<34.Tn<34.導(dǎo)學(xué)號168041927.(2017湖南長郡中學(xué)模擬,理17)在數(shù)列an中,Sn為其前n項(xiàng)和,an>0,且4Sn=an2+2an+1(nN*),數(shù)列bn為等比數(shù)列,公比q>1,b1=a1,且2b2,b4,3b3成等差數(shù)列.(1)求an與bn的通項(xiàng)公式;(2)令cn=anbn,若cn的前n項(xiàng)和為Tn,求證:Tn<6.(1)解 4Sn=an2+2an+1(nN*),當(dāng)n=1時(shí),4a1=a12+2a1+1,解得a1=1.當(dāng)n2時(shí),4Sn-1=an-12+2an-1+1,-得4an=(an+1)2-(an-1+1)2,即(an+an-1)(an-an-1-2)=0.又an>0,an-an-1-2=0,即an-an-1=2,數(shù)列an是等差數(shù)列,公差為2.an=1+2(n-1)=2n-1.2b2,b4,3b3成等差數(shù)列,2b4=2b2+3b3.2b2q2=2b2+3b2q,即2q2-3q-2=0,解得q=2.又b1=a1=1,bn=2n-1.(2)證明 cn=anbn=2n-12n-1,則cn的前n項(xiàng)和為Tn=1+32+522+2n-12n-1,12Tn=12+322+2n-32n-1+2n-12n,12Tn=1+212+122+12n-1-2n-12n=1+2×121-12n-11-12-2n-12n,Tn=6-2n+32n-1<6.導(dǎo)學(xué)號168041938.已知數(shù)列an的前n項(xiàng)和為Sn,a1=1,且對任意正整數(shù)n,點(diǎn)(an+1,Sn)都在直線2x+y-2=0上.(1)求數(shù)列an的通項(xiàng)公式;(2)是否存在實(shí)數(shù),使得數(shù)列Sn+n+2n為等差數(shù)列?若存在,求出的值;若不存在,請說明理由.解 (1)由題意,得2an+1+Sn-2=0.當(dāng)n2時(shí),2an+Sn-1-2=0.-,得2an+1-2an+an=0(n2),所以an+1an=12(n2).因?yàn)閍1=1,2a2+a1=2,所以a2=12.所以an是首項(xiàng)為1,公比為12的等比數(shù)列.所以數(shù)列an的通項(xiàng)公式為an=12n-1.(2)由(1)知,Sn=1-12n1-12=2-12n-1.若Sn+n+2n為等差數(shù)列,則S1+2,S2+2+22,S3+3+23成等差數(shù)列,則2S2+94=S1+32+S3+258,即232+94=1+32+74+258,解得=2.又當(dāng)=2時(shí),Sn+2n+22n=2n+2,顯然2n+2是等差數(shù)列.故存在實(shí)數(shù)=2,使得數(shù)列Sn+n+2n為等差數(shù)列.導(dǎo)學(xué)號16804194