2025高考数学考二轮专专题突破练12求数列的通项及前n项和-专项训练【含答案】

这是一份2025高考数学考二轮专专题突破练12求数列的通项及前n项和-专项训练【含答案】，共9页。
(1)求数列{an}的通项公式;
(2)设bn=lg3a1+lg3a2+…+lg3an,求数列1bn的前n项和Tn.
2.已知等差数列{an}的前n项和为Sn,且满足a3=8,S5=2a7.
(1)求数列{an}的通项公式;
(2)若数列{bn}满足bn=ancs nπ+2n+1,求数列{bn}的前2n项和T2n.
3.(2024·山西吕梁二模)已知正项数列{an}的前n项和为Sn,且满足6Sn=(an+2)(an+1),首项a1≠1,n∈N*.
(1)求数列{an}的通项公式;
(2)证明:1S1+1+1S2+2+…+1Sn+n5时,Tn>Sn.
5.已知等差数列{an}与正项等比数列{bn}满足a1=b1=3,且b3-a3,20,a5+b2既是等差数列,又是等比数列.
(1)求数列{an}和{bn}的通项公式;
(2)在以下三个条件中任选一个,补充在下面问题中,并完成求解.
①cn=1an·an+1+(-1)nbn,②cn=an·bn,③cn=2(an+3)anan+1bn+1.
若 ,求数列{cn}的前n项和Sn.
6.已知等比数列{an}的前n项和为Sn,且an+1=2Sn+2,数列{bn}满足b1=2,(n+2)bn=nbn+1.
(1)求数列{an}和{bn}的通项公式;
(2)在an与an+1之间插入n个数,使这n+2个数组成一个公差为cn的等差数列,求数列{bncn}的前n项和Tn.
7.已知数列{an}满足a1=1,an+1=3an+3n+1.
(1)求证:数列an3n是等差数列;
(2)求数列{an}的通项公式;
(3)设数列{an}的前n项和为Sn,求证:Sn3n>3n2−74.
8.(2023·新高考Ⅰ,20)设等差数列{an}的公差为d,且d>1.令bn=n2+nan,记Sn,Tn分别为数列{an},{bn}的前n项和.
(1)若3a2=3a1+a3,S3+T3=21,求数列{an}的通项公式;
(2)若{bn}为等差数列,且S99-T99=99,求d.
专题突破练12 求数列的通项及前n项和 答案
1.解 (1)设等比数列{an}的公比为q(q>0),由a32=9a2a6,得a32=9a42,所以q2=19,所以q=13.
由2a1+3a2=1,得2a1+3a1·13=1,
所以a1=13.
故数列{an}的通项公式为an=13n.
(2)因为bn=lg3a1+lg3a2+…+lg3an=-(1+2+…+n)=-n(n+1)2,
所以1bn=-2n(n+1)=-21n-1n+1.
所以Tn=1b1+1b2+…+1bn=-2[1-12+12-13+…+1n-1n+1]=-2nn+1.
所以数列1bn的前n项和Tn=-2nn+1.
2.解 (1)设{an}的公差为d,依题意,a1+2d=8,5a1+5×42d=2(a1+6d),
解得a1=2,d=3.所以an=2+3(n-1)=3n-1.
(2)因为bn=ancs nπ+2n+1=(-1)nan+2n+1=(-1)n·(3n-1)+2n+1,
所以T2n=(a2-a1)+(a4-a3)+…+(a2n-a2n-1)+(22+23+…+22n+1)=3n+22(1-22n)1-2=3n+22n+2-4.
3.(1)解 由6Sn=(an+2)(an+1),可得6Sn+1=(an+1+2)(an+1+1),
两式相减可得6an+1=(an+1+1)(an+1+2)-(an+1)(an+2),
化简可得(an+1+an)(an+1-an-3)=0,
由正项数列{an}知an+1+an>0,所以an+1-an=3,
又6S1=(a1+2)(a1+1),解得a1=2或a1=1(舍去),
所以{an}是以2为首项,3为公差的等差数列,
故an=2+3(n-1)=3n-1.
(2)证明 由(1)可得Sn=n(a1+an)2=n(2+3n-1)2=3n2+n2,
所以1Sn+n=13n2+3n2=23n(n+1)=23(1n−1n+1),
因此1S1+1+1S2+2+…+1Sn+n=23(1-12+12−13+…+1n−1n+1)=23×(1-1n+1)5时,Tn-Sn=3n2+5n-102-(n2+4n)=n2-3n-102=(n-5)(n+2)2>0,所以Tn>Sn.
当n为偶数时,Tn=a1-6+2a2+a3-6+2a4+a5-6+2a6+…+an-1-6+2an=(-1+14)+(3+22)+(7+30)+…+[(2n-5)+(4n+6)]=[-1+3+…+(2n-5)]+[14+22+…+(4n+6)]=n2(-1+2n-5)2+n2(14+4n+6)2=3n2+7n2.
当n>5时,Tn-Sn=3n2+7n2-(n2+4n)=n2-n2=n(n-1)2>0,所以Tn>Sn.
综上可知,当n>5时,Tn>Sn.
5.解 (1)设等差数列{an}的公差为d,等比数列{bn}的公比为q(q>0),由已知得20=b3-a3=a5+b2,即20=3q2-(3+2d),20=(3+4d)+3q,解得d=2,q=3,所以an=2n+1,bn=3n.
(2)若选择①,
则cn=1an·an+1+(-1)nbn=1(2n+1)(2n+3)+(-3)n=1212n+1-12n+3+(-3)n,
所以Sn=c1+c2+…+cn=12×13-15+(-3)1+12×15-17+(-3)2+…+12(12n+1−12n+3)+(-3)n=1213-12n+3+-3[1-(-3)n]1+3=n3(2n+3)−3[1-(-3)n]4.
若选择②,
则cn=an·bn=(2n+1)3n,
所以Sn=c1+c2+…+cn=3×3+5×32+…+(2n+1)3n,3Sn=3×32+5×33+…+(2n+1)3n+1,
两式相减得-2Sn=32+2×32+2×33+…+2×3n-(2n+1)3n+1=-2n·3n+1,所以Sn=n·3n+1.
若选择③,
则cn=2(an+3)anan+1bn+1=2(2n+4)(2n+1)(2n+3)3n+1=1(2n+1)3n−1(2n+3)3n+1,
所以Sn=c1+c2+…+cn=13×3-15×32+(15×32−17×33)+…+[1(2n+1)3n−1(2n+3)3n+1]=19−1(2n+3)3n+1.
6.解 (1)设等比数列{an}的公比为q,由an+1=2Sn+2,可得an=2Sn-1+2(n≥2),
两式相减得an+1-an=2Sn-2Sn-1=2an,
整理得an+1=3an,可知q=3.
令n=1,则a2=2a1+2,即3a1=2a1+2,解得a1=2.
故an=2·3n-1.
由b1=2,(n+2)bn=nbn+1,得bn+1bn=n+2n,
则当n≥2时,bn=bnbn-1·bn-1bn-2·…·b2b1·b1=n+1n-1·nn-2·…·31×2=n(n+1).
又b1=2满足上式,所以bn=n(n+1).
(2)若在an与an+1之间插入n个数,使这n+2个数组成一个公差为cn的等差数列,
则an+1-an=(n+1)cn,即2·3n-2·3n-1=(n+1)cn,整理得cn=4·3n-1n+1,所以bncn=4n·3n-1,
所以Tn=b1c1+b2c2+b3c3+…+bn-1cn-1+bncn=4×1×30+4×2×31+4×3×32+…+4·(n-1)3n-2+4·n·3n-1=4[1×30+2×31+3×32+…+(n-1)3n-2+n·3n-1],
3Tn=4[1×31+2×32+…+(n-1)3n-1+n·3n],
两式相减得-2Tn=4(30+31+32+…+3n-1-n·3n)=41-3n1-3-n·3n=(2-4n)·3n-2,所以Tn=(2n-1)3n+1.
7.(1)证明 由an+1=3an+3n+1,得an+13n+1=an3n+1,即an+13n+1−an3n=1.
又a13=13,所以数列an3n是以13为首项,1为公差的等差数列.
(2)解 由(1)得an3n=13+(n-1)×1=n-23,
所以an=n-23·3n.
(3)证明 由(2)得Sn=1-23×31+2-23×32+…+(n-1)-23×3n-1+n-23×3n,
3Sn=1-23×32+2-23×33+…+(n-1)-23×3n+n-23×3n+1,
两式相减得2Sn=n-23×3n+1-3n+1-92-1=n-76×3n+1+72,
故Sn=n2-7123n+1+74,从而Sn3n=n2-7123n+13n+74×3n=3n2-74+74×3n>3n2−74.
8.解 (1)由3a2=3a1+a3,得3(a2-a1)=a3,
即3d=a1+2d,得a1=d,
从而an=nd,故bn=n2+nnd=n+1d.
易知S3=a1+a2+a3=6d,T3=2+3+4d=9d.
由题意得6d+9d=21,从而2d2-7d+3=0.
整理得(2d-1)(d-3)=0,解得d=3或d=12(舍去).故an=3n.
(2)由题意,a2=a1+d,a3=a1+2d,b1=2a1,b2=6a2,b3=12a3,∵{bn}为等差数列,
∴2b2=b1+b3,即2×6a2=2a1+12a3,
∴2×6a1+d=2a1+12a1+2d,
解得a1=d或a1=2d.
当a1=d时,an=a1+(n-1)d=d+d(n-1)=nd,bn=n2+nnd=n+1d=2d+1d(n-1),
此时{bn}是以2d为首项,1d为公差的等差数列,
S99=99(a1+a99)2=99(2a1+98d)2=99(2d+98d)2=99×50d,
T99=99(b1+b99)2=99(2b1+98×1d)2=99(2×2d+98×1d)2=99×51d.
S99-T99=99×50d-99×51d=99,
解得d=5150或d=-1(舍去).
当a1=2d时,an=a1+(n-1)d=2d+d(n-1)=(n+1)d,
bn=n2+n(n+1)d=nd=1d+1d(n-1),此时{bn}是以1d为首项,1d为公差的等差数列,
S99=99(a1+a99)2=99(2a1+98d)2=99(4d+98d)2=99×51d,
T99=99(b1+b99)2=99(2b1+98×1d)2=99(2×1d+98×1d)2=99×50d,
S99-T99=99×51d-99×50d=99,
解得d=-5051