（广西课标版）2020版高考数学二轮复习题型练4大题专项2文.docx

题型练4大题专项(二)数列的通项、求和问题1.(2019湖北4月调研,17)已知数列an满足a2-a1=1,其前n项和为Sn,当n2时,Sn-1-1,Sn,Sn+1成等差数列.(1)求证an为等差数列;(2)若Sn=0,Sn+1=4,求n.2.(2019吉林实验中学检测,17)已知等差数列an满足a4=7,2a3+a5=19.(1)求an;(2)设bn-an是首项为2,公比为2的等比数列,求数列bn的通项公式及前n项和Tn.3.设an是等差数列,且a1=ln 2,a2+a3=5ln 2.(1)求an的通项公式.(2)求ea1+ea2+ean.4.已知等差数列an的前n项和为Sn,公比为q的等比数列bn的首项是12,且a1+2q=3,a2+4b2=6,S5=40.(1)求数列an,bn的通项公式an,bn;(2)求数列1anan+1+1bnbn+1的前n项和Tn.5.已知函数f(x)=7x+5x+1,数列an满足:2an+1-2an+an+1an=0,且anan+10.在数列bn中,b1=f(0),且bn=f(an-1).(1)求证:数列1an是等差数列;(2)求数列|bn|的前n项和Tn.6.已知等比数列an的公比q1,且a3+a4+a5=28,a4+2是a3,a5的等差中项.数列bn满足b1=1,数列(bn+1-bn)an的前n项和为2n2+n.(1)求q的值;(2)求数列bn的通项公式.题型练4大题专项(二)数列的通项、求和问题1.(1)证明当n2时,由Sn-1-1,Sn,Sn+1成等差数列,可知2Sn=Sn-1-1+Sn+1,即Sn-Sn-1=-1+Sn+1-Sn,即an=-1+an+1(n2),则an+1-an=1(n2),又a2-a1=1,故an是公差为1的等差数列.(2)解由(1)知等差数列an的公差为1.由Sn=0,Sn+1=4,得an+1=4,即a1+n=4.由Sn=0,得na1+n(n-1)2=0,即a1+n-12=0,解得n=7.2.解(1)由题意得a1+3d=7,2(a1+2d)+a1+4d=19,解得a1=1,d=2.an=1+2(n-1)=2n-1.(2)由题意可知bn-an=2n,bn=2n+2n-1,Tn=(2+22+2n)+1+3+(2n-1),Tn=2n+1+n2-2.3.解(1)设等差数列an的公差为d,a2+a3=5ln2.2a1+3d=5ln2,又a1=ln2,d=ln2.an=a1+(n-1)d=nln2.(2)由(1)知an=nln2.ean=enln2=eln2n=2n,ean是以2为首项,2为公比的等比数列.ea1+ea2+ean=2+22+2n=2n+1-2.ea1+ea2+ean=2n+1-2.4.解(1)设an公差为d,由题意得a1+2d=8,a1+2q=3,a1+d+2q=6,解得a1=2,d=3,q=12,故an=3n-1,bn=12n.(2)1anan+1+1bnbn+1=131an-1an+1+1bnbn+1=131an-1an+1+22n+1,Tn=1312-15+15-18+13n-1-13n+2+8(1-4n)1-4=1312-13n+2+13(22n+3-8)=1322n+3-13n+2-52.5.(1)证明2an+1-2an+an+1an=0,1an+1-1an=12,故数列1an是以12为公差的等差数列.(2)解b1=f(0)=5,7(a1-1)+5a1-1+1=5,7a1-2=5a1,a1=1,1an=1+(n-1)12,an=2n+1,bn=7an-2an=7-(n+1)=6-n.当n6时,Tn=n2(5+6-n)=n(11-n)2;当n7时,Tn=15+n-62(1+n-6)=n2-11n+602.故Tn=n(11-n)2,n6,n2-11n+602,n7.6.解(1)由a4+2是a3,a5的等差中项,得a3+a5=2a4+4,所以a3+a4+a5=3a4+4=28,解得a4=8.由a3+a5=20,得8q+1q=20,解得q=2或q=12,因为q1,所以q=2.(2)设cn=(bn+1-bn)an,数列cn前n项和为Sn,由cn=S1,n=1,Sn-Sn-1,n2,解得cn=4n-1.由(1)可知an=2n-1,所以bn+1-bn=(4n-1)12n-1.故bn-bn-1=(4n-5)12n-2,n2,bn-b1=(bn-bn-1)+(bn-1-bn-2)+(b3-b2)+(b2-b1)=(4n-5)12n-2+(4n-9)12n-3+712+3.设Tn=3+712+11122+(4n-5)12n-2,n2,12Tn=312+7122+(4n-9)12