高考数学二轮复习 专题对点练14 数列与数列不等式的证明及数列中的存在性问题 理.doc

专题对点练14数列与数列不等式的证明及数列中的存在性问题1.若数列an满足:a1=23,a2=2,3(an+1-2an+an-1)=2.(1)证明:数列an+1-an是等差数列;(2)求使1a1+1a2+1a3+1an52成立的最小的正整数n.(1)证明 由3(an+1-2an+an-1)=2可得an+1-2an+an-1=23,即(an+1-an)-(an-an-1)=23,故数列an+1-an是以a2-a1=43为首项,23为公差的等差数列.(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+152,n5.最小的正整数n为6.2.(2017广东揭阳二模,理17)已知数列an,a1=1,an+1=2(n+1)ann+n+1.(1)求证:数列ann+1是等比教列;(2)求数列an的前n项和sn.(1)证明 an+1=2(n+1)ann+n+1,an+1n+1=2ann+1,an+1n+1+1=2ann+1,数列ann+1是等比教列,公比为2,首项为2.(2)解 由(1)可得ann+1=2n,可得an=n2n-n.设数列n2n的前n项和为tn,则tn=2+222+323+n2n,2tn=22+223+(n-1)2n+n2n+1,两式相减可得-tn=2+22+2n-n2n+1=2(1-2n)1-2-n2n+1,tn=(n-1)2n+1+2.sn=(n-1)2n+1+2-n(n+1)2.3.已知数列an的前n项和sn=1+an,其中0.(1)证明an是等比数列,并求其通项公式;(2)若s5=3132,求.解 (1)由题意得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是首项为11-,公比为-1的等比数列,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)在数列an中,设f(n)=an,且f(n)满足f(n+1)-2f(n)=2n(nn*),且a1=1.(1)设bn=an2n-1,证明数列bn为等差数列;(2)求数列an的前n项和sn.(1)证明 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是首项为1,公差为1的等差数列.(2)解 由(1)知,bn=an2n-1=n,an=n2n-1.sn=1+221+322+n2n-1,2sn=121+222+(n-1)2n-1+n2n,两式相减得-sn=1+21+22+2n-1-n2n=2n-1-n2n=(1-n)2n-1,sn=(n-1)2n+1.5.设数列an的前n项和为sn,且(3-m)sn+2man=m+3(nn*),其中m为常数,且m-3.(1)求证:an是等比数列;(2)若数列an的公比q=f(m),数列bn满足b1=a1,bn=32f(bn-1)(nn*,n2),求证:1bn为等差数列,并求bn.证明 (1)由(3-m)sn+2man=m+3,得(3-m)sn+1+2man+1=m+3,两式相减,得(3+m)an+1=2man.m-3,an+1an=2mm+3,an是等比数列.(2)由(3-m)sn+2man=m+3,得(3-m)s1+2ma1=m+3,即a1=1,b1=1.数列an的公比q=f(m)=2mm+3,当n2时,bn=32f(bn-1)=322bn-1bn-1+3,bnbn-1+3bn=3bn-1,1bn-1bn-1=13.1bn是以1为首项,13为公差的等差数列,1bn=1+n-13=n+23.又1b1=1也符合,bn=3n+2.6.已知数列an的前n项和为sn,a1=-2,且满足sn=12an+1+n+1(nn*).(1)求数列an的通项公式;(2)若bn=log3(-an+1),求数列1bnbn+2的前n项和tn,并求证tn34.(1)解 sn=12an+1+n+1(nn*),当n=1时,-2=12a2+2,解得a2=-8.当n2时,an=sn-sn-1=12an+1+n+1-12an+n,即an+1=3an-2,可得an+1-1=3(an-1).当n=1时,a2-1=3(a1-1)=-9,数列an-1是等比数列,首项为-3,公比为3.an-1=-3n,即an=1-3n.(2)证明 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+234.tn0,且4sn=an2+2an+1(nn*),数列bn为等比数列,公比q1,b1=a1,且2b2,b4,3b3成等差数列.(1)求an与bn的通项公式;(2)令cn=anbn,若cn的前n项和为tn,求证:tn0,an-an-1-2=0,即an-an-1=2,数列an是等差数列,公差为2.an=1+2(n-1)=2n-1.2b2,b4,3b3成等差数列,2b4=2b2+3b3.2b2q2=2b2+3b2q,即2q2-3q-2=0,解得q=2.又b1=a1=1,bn=2n-1.(2)证明 cn=anbn=2n-12n-1,则cn的前n项和为tn=1+32+522+2n-12n-1,12tn=12+322+2n-32n-1+2n-12n,12tn=1+212+122+12n-1-2n-12n=1+2121-12n-11-12-2n-12n,tn=6-2n+32n-16.导学号168041938.已知数列an的前n项和为sn,a1=1,且对任意正整数n,点(an+1,sn)都在直线2x+y-2=0上.(1)求数列an的通项公式;(2)是否存在实数,使得数列sn+n+2n为等差数列?若存在,求出的值;若不存在,请说明理由.解 (1)由题意,得2an+1+sn-2=0.当n2时,2an+sn-1-2=0.-,得2an+1-2an+an=0(n2),所以an+1an=12(n2).因为a1=1,2a2+a1=2,所以a2=12.所以an是首项为1,公比为12的等比数列.所以数列an的通项公式为an=12n-1.(2)由(1)知,sn=1-12n1-12=2-12n-1.若s