2020届二轮（理科数学） 等差、等比数列的综合问题 专题卷（全国通用）

专题突破练13　等差、等比数列的综合问题1.(2019黑龙江哈尔滨第三中学高三第二次模拟)设数列{an}满足an+1=an+2,a1=4.(1)求证:{an-3}是等比数列,并求an;(2)求数列{an}的前n项和Tn.              2.(2019湖北高三4月份调研)已知数列{an}满足a2-a1=1,其前n项和为Sn,当n≥2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. 3.(2019贵州贵阳高三5月适应性考试)等差数列{an}的前n项和为Sn,公差d≠0,已知S4=16,a1,a2,a5成等比数列.(1)求数列{an}的通项公式;(2)记点A(n,Sn),B(n+1,Sn+1),C(n+2,Sn+2),求证:△ABC的面积为1.              4.已知等比数列{an}的前n项和为Sn,a1=3,且3S1,2S2,S3成等差数列.(1)求数列{an}的通项公式;(2)设bn=log3an,求Tn=b1b2-b2b3+b3b4-b4b5+…+b2n-1b2n-b2nb2n+1.                       5.(2019广东梅州高三总复习质检)已知数列{an}满足(a1+2a2+…+2n-1an)=2n+1(n∈N*).(1)求a1,a2和{an}的通项公式;(2)记数列{an-kn}的前n项和为Sn,若Sn≤S4对任意的正整数n恒成立,求实数k的取值范围.              6.(2019西藏山南地区第二高级中学高三上学期期中模拟)已知等差数列{an}的前n项和为Sn,且S9=90,S15=240.(1)求数列{an}的通项公式和前n项和Sn;(2)设{bn-(-1)nan}是等比数列,且b2=7,b5=71,求数列{bn}的前n项和Tn. 7.(2019山东烟台高三5月适应性练习)已知数列{an}前n项和Sn满足Sn=2an-2(n∈N*),{bn}是等差数列,且a3=b4-2b1,b6=a4.(1)求{an}和{bn}的通项公式;(2)求数列{(-1)n}的前2n项和T2n.              8.设{an}是等差数列,其前n项和为Sn(n∈N*);{bn}是等比数列,公比大于0,其前n项和为Tn(n∈N*).已知b1=1,b3=b2+2,b4=a3+a5,b5=a4+2a6.(1)求Sn和Tn;(2)若Sn+(T1+T2+…+Tn)=an+4bn,求正整数n的值.  参考答案 专题突破练13　等差、等比数列的综合问题1.解(1)∵an+1=an+2,a1=4,∴an+1-3=(an-3).故{an-3}是首项为1,公比为的等比数列.∴an=3+n-1.(2)an=3+n-1,故Tn=3n+0+1+…+n-1=3n+=3n+1-n.2.(1)证明当n≥2时,由Sn-1-1,Sn,Sn+1成等差数列,得2Sn=Sn-1-1+Sn+1,即Sn-Sn-1=-1+Sn+1-Sn,即an=-1+an+1(n≥2),则an+1-an=1(n≥2),又a2-a1=1,故{an}是公差为1的等差数列.(2)解由(1)知数列{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+=0,即a1+=0,联立解得n=7.3.(1)解由题意得由于d≠0,解得∴an=1+(n-1)×2=2n-1.(2)证明由(1)知Sn=n×1+2=n2,∴△ABC的面积S=(Sn+Sn+2)×2-(Sn+Sn+1)×1-(Sn+1+Sn+2)×1=(Sn+Sn+2-2Sn+1)=[n2+(n+2)2-2(n+1)2]=1.4.解(1)∵3S1,2S2,S3成等差数列,∴4S2=3S1+S3,∴4(a1+a2)=3a1+(a1+a2+a3),即a3=3a2,∴公比q=3,∴an=a1qn-1=3n.(2)由(1)知,bn=log3an=log33n=n,∵b2n-1b2n-b2nb2n+1=(2n-1)·2n-2n(2n+1)=-4n,∴Tn=(b1b2-b2b3)+(b3b4-b4b5)+…+(b2n-1b2n-b2nb2n+1)=-4(1+2+…+n)=-4=-2n2-2n.5.解(1)由题意得a1+2a2+…+2n-1an=n·2n+1,所以a1=1×22=4,a1+2a2=2×23,得a2=6.由a1+2a2+…+2n-1an=n·2n+1,所以a1+2a2+…+2n-2an-1=(n-1)·2n(n≥2),相减得2n-1an=n·2n+1-(n-1)·2n,得an=2n+2,当n=1也满足上式.所以{an}的通项公式为an=2n+2.(2)数列{an-kn}的通项公式为an-kn=2n+2-kn=(2-k)n+2,所以数列{an-kn}是以4-k为首项,公差为2-k的等差数列.若Sn≤S4对任意的正整数n恒成立,等价于当n=4时,Sn取得最大值,所以解得k6.解(1)设等差数列{an}的首项为a1,公差为d.则由得解得所以an=2+(n-1)×2=2n,即an=2n.Sn=2n+2=n(n+1),即Sn=n(n+1).(2)令cn=bn-(-1)nan,设{cn}的公比为q,∵b2=7,b5=71,an=2n,∴c2=b2-(-1)2a2=3,c5=b5-(-1)5a5=81,∴q3==27,q=3,∴cn=c2qn-2=3n-1,从而bn=3n-1+(-1)n2n,Tn=b1+b2+…+bn=(30+31+…+3n-1)+[-2+4-6+…+(-1)n2n],当n为偶数时,Tn=;当n为奇数时,Tn=所以Tn= 7.解(1)Sn=2an-2,①当n=1时,得a1=2,当n≥2时,Sn-1=2an-1-2,②①②两式作差得an=2an-1(n≥2),所以数列{an}是以2为首项,公比为2的等比数列,所以an=2n.设等差数列{bn}的公差为d,由所以所以所以bn=3n-2.(2)T2n=(-)+(-)+…+(-)=3(b1+b2)+3(b3+b4)+…+3(b2n-1+b2n)=3(b1+b2)+3(b3+b4)+…+3(b2n-1+b2n)=3(b1+b2+…+b2n).又因为bn=3n-2,所以T2n=3=3n[1+3×(2n)-2]=18n2-3n.8.解(1)设等比数列{bn}的公比为q.由b1=1,b3=b2+2,可得q2-q-2=0.因为q>0,可得q=2,故bn=2n-1.所以,Tn==2n-1.设等差数列{an}的公差为d.由b4=a3+a5,可得a1+3d=4.由b5=a4+2a6,可得3a1+13d=16,从而a1=1,d=1,故an=n.所以Sn=(2)由(1),有T1+T2+…+Tn=(21+22+…+2n)-n=-n=2n+1-n-2.由Sn+(T1+T2+…+Tn)=an+4bn可得+2n+1-n-2=n+2n+1,整理得n2-3n-4=0,解得n=-1(舍),或n=4.所以正整数n的值为4.