1、专题突破练13等差、等比数列的综合问题1.(2023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=13an+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.(2023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.3.(2023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d0,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
2、,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.(2023广东梅州高三总复习质检)已知数列an满足1n(a1+2a2+2n-1an)=2n+1(nN*).(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SnS4对任意的正整数n恒成立,求实数k的取值范围.6.(2023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
3、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.(2023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(nN*),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)nbn2的前2n项和T2n.8.设an是等差数列,其前n项和为Sn(nN*);bn是等比数列,公比大于0,其前n项和为Tn(nN*).已知b1=1,b3=b2+2,b4=a3+a5,b5=a4+2a6.(1)求Sn和Tn;(
4、2)若Sn+(T1+T2+Tn)=an+4bn,求正整数n的值.2023学年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=13an+2,a1=4,an+1-3=13(an-3).故an-3是首项为1,公比为13的等比数列.an=3+13n-1.(2)an=3+13n-1,故Tn=3n+130+131+13n-1=3n+1-(13)n1-13=3n+321-13n.2.(1)证明当n2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(n2),则an+1-an=1(n2),又a2-a
5、1=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+n(n-1)2=0,即a1+n-12=0,联立解得n=7.3.(1)解由题意得4a1+432d=16,(a1+d)2=a1(a1+4d),由于d0,解得a1=1,d=2.an=1+(n-1)2=2n-1.(2)证明由(1)知Sn=n1+n(n-1)22=n2,ABC的面积S=12(Sn+Sn+2)2-12(Sn+Sn+1)1-12(Sn+1+Sn+2)1=12(Sn+Sn+2-2Sn+1)=12n2+(n+2)2-2(n+1)2=1.4.解(
6、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)=-4n(n+1)2=-2n2-2n.5.解(1)由题意得a1+2a2+2n-1an=n2n+1,所以a1=122=4,a1+2a2=223,得a2=6.由a1+2a2+2n-1an=n2n+
7、1,所以a1+2a2+2n-2an-1=(n-1)2n(n2),相减得2n-1an=n2n+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SnS4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a4-4k=4(2-k)+20,a5-5k=5(2-k)+20.解得125k52.6.解(1)设等差数列an的首项为a1,公差为d.则由S9=90,S15=240,得9a1+36d=90,15a1+105
8、d=240,解得a1=2,d=2.所以an=2+(n-1)2=2n,即an=2n.Sn=2n+n(n-1)22=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=c5c2=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=3n+2n-12;当n为奇数时,Tn=3n-2n-32.所以Tn=3n+2n-12,n为偶数,3n-2
9、n-32,n为奇数.7.解(1)Sn=2an-2,当n=1时,得a1=2,当n2时,Sn-1=2an-1-2,两式作差得an=2an-1(n2),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a3=b4-2b1,b6=a4,所以8=3d-b1,16=5d+b1,所以d=3,b1=1,所以bn=3n-2.(2)T2n=(-b12+b22)+(-b32+b42)+(-b2n-12+b2n2)=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=3
10、n-2,所以T2n=32n(b1+b2n)2=3n1+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q0,可得q=2,故bn=2n-1.所以,Tn=1-2n1-2=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=n(n+1)2.(2)由(1),有T1+T2+Tn=(21+22+2n)-n=2(1-2n)1-2-n=2n+1-n-2.由Sn+(T1+T2+Tn)=an+4bn可得n(n+1)2+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4.12