全品高考备战2027年数学一轮学生用书05第40讲数列的综合问题【答案】作业手册

第40讲数列的综合问题1.C[解析]设等差数列{an}的公差为d(d≠0).因为a3,a4,a6成等比数列,且a1=-2,所以a42=a3a6,即(-2+3d)2=(-2+2d)(-2+5d),解得d=2或d=0(舍去),所以a10=a1+9d=-2+9×2=16.2.A[解析]∵2S3,3S5,4S6成等差数列,∴6S5=2S3+4S6,即6(a1+a2+a3+a4+a5)=2(a1+a2+a3)+4(a1+a2+a3+a4+a5+a6),整理得6a4+6a5=4a4+4a5+4a6,即2a6-a5-a4=0,∵a4≠0,∴2q2-q-1=0,解得q=1或q=-12,故选A3.C[解析]因为{an}既是等差数列又是等比数列,且a1=1,所以an=1(n∈N*)(常数数列),所以其前2026项和为2026,故选C.4.D[解析]设数列{an}的公差为d,因为a7=2,a9,a5,a13成等比数列,所以a1+6d=2,(a1+8d)(a1+12d)=(a1+4d)2,解得a1=20,d=-3,5.A[解析]数列{an}满足an=n,在an,an+1之间插入n个1,构成数列{bn}:a1,1,a2,1,1,a3,1,1,1,a4,…,所以数列{bn}中从a1到ak的项数为k+[1+2+…+(k-1)]=k+(k-1)(1+k-1)2=12k(k+1),当k=13时,12×13×14=91,当k=14时,12×14×15=105,因为ak=k,所以S100=(a6.C[解析]令an=n-32n-17≥0,解得n≤3或n>172,所以当n≤3时,an≥0,当4≤n≤8时,an<0,当n≥9时,an>0,且a1=215,a2=113,a3=0,a4=-19,…,a8=-5,所以S8<S7.2和3[解析]an=nn2+6=1n+6n,∵y=x+6x在(0,6)上单调递减,在(6,+∞)上单调递增,且a2=8.107,2[解析]因为Sn=an2-10n,且对任意的n∈N*,都有S3≤Sn,所以52≤--102a=59.解:(1)设{an}的公差为d,则a3=a1+2d=7,S6=6a1+15d=48,解得a1=3,d=2,所以an=3+(n-1)×2=2n+1.(2)证明:因为1a1(2n+1)(2n+5)=14×12n+1-12n+5,所以因为n∈N*,所以n+2(2n+3)(2所以1a1a3+1a10.D[解析]因为b为a和2的等差中项,所以a+22=b,即a=2b-2,则3a+19b=32b-2+132b≥232b-2·132b=11.A[解析]由an+1=3an+2,可得an+1+1=3(an+1),则{an+1}是首项为3,公比为3的等比数列,所以an+1=3n,即an=3n-1,则an+1(12×13n+2-13n+1+2,所以Tn=12×15-111+111-129+…12.ACD[解析]当n=1时,S1=2a1-1,即a1=2a1-1,所以a1=1,当n≥2时,Sn-1=2an-1-1,所以Sn-Sn-1=(2an-1)-(2an-1-1),即an=2an-2an-1,即an=2an-1,因为a1=1≠0,所以anan-1=2(n≥2),所以{an}是首项为1,公比为2的等比数列,所以an=2n-1,故A正确;由(n+2)bn=nbn+1,得bnn=bn+1n+2,则bn(n+1)n=bn+1(n+2)(n+1),所以数列bn(n+1)n为常数列,所以bn(n+1)n=b12×1=1,所以bn=n(n+1)=n2+n,故B错误;cn=23n-1·(n2+n),当n≥2时,cncn-1=2(n2+n)3(n2-n)=2(n+1)3(n-13.6[解析]因为an+1=an2+2an,所以an+1+1=(an+1)2,则ln(an+1+1)=2ln(an+1),又a1+1=2,所以数列{ln(an+1)}是首项为ln2,公比为2的等比数列,所以ln(an+1)=2n-1ln2,所以an=22n-1-1.由an+1>2025,得22n-12>2025,因为210=1024,21114.an+1=an+algan-ban2-can(n∈N*)[解析]因为第n年鱼类的自然繁殖量与lgan成正比,自然死亡量与an2成正比,捕捞量与an成正比,比例系数分别为a,b,c(a,b,c>0),所以an+1=an+algan-ban2-can(n∈N*).若b=1,a1=10,要保持每年年初时鱼类总量始终不变,即an+1=an,则algan-an2-can=0,代入n=1,得a-10c-100=0,即a=10c+100,由a,c>0,不妨取c=1,a15.解:(1)∵4Sn+1=3Sn-9①,∴当n≥2时,4Sn=3Sn-1-9②,由①-②得4an+1=3an(n≥2),在①式中,令n=1,则4-94+a2=-274-9,∴a2=-2716,满足a2a1=34,∴an+1an=34对n∈N*恒成立,∴{an(2)由3bn+(n-4)an=0,得bn=(n-4)·34n,∴Tn=(-3)×34+(-2)×342+(-1)×343+…+(n-5)·34n-1+(n-4)·34n③,∴34Tn=(-3)×342+(-2)×343+…+(n-6)·34n-1+(n-5)·34n+(n-4)·34n+1④,由③-④得14∴Tn=-4n·34由Tn≤λbn得-4n·34n+1≤λ·(n-4)·34n,即(当1≤n<4时,可得λ≤-3此时λ≤-3当n=4时,不等式显然成立;当n>4时,可得λ≥-3nn-4,∵-∴λ≥-3.综上,-3≤λ≤1.16.解:(1)由bn+1=3bn-2n+1,得bn+1-(n+1)=3(bn-n),又b1-1=3,所以{bn-n}是首项为3,公比为3的等比数列,所以bn