《（通用版）2020版高考數(shù)學(xué)大二輪復(fù)習(xí) 大題專項(xiàng)練（二）數(shù)列 理》由會(huì)員分享，可在線閱讀，更多相關(guān)《（通用版）2020版高考數(shù)學(xué)大二輪復(fù)習(xí) 大題專項(xiàng)練（二）數(shù)列 理（8頁(yè)珍藏版）》請(qǐng)?jiān)谘b配圖網(wǎng)上搜索。

1、大題專項(xiàng)練(二)　數(shù)列 A組　基礎(chǔ)通關(guān) 1.已知等差數(shù)列{an}滿足a3-a2=3,a2+a4=14. (1)求{an}的通項(xiàng)公式; (2)設(shè)Sn是等比數(shù)列{bn}的前n項(xiàng)和,若b2=a2,b4=a6,求S7. 解(1)設(shè)等差數(shù)列{an}的公差為d, ∵a3-a2=3,a2+a4=14. ∴d=3,2a1+4d=14, 解得a1=1,d=3, ∴an=1+3(n-1)=3n-2. (2)設(shè)等比數(shù)列{bn}的公比為q,b2=a2=4=b1q,b4=a6=16=b1q3,聯(lián)立解得b1=2,q=2,或b1=-2,q=-2. ∴S7=2×(27-1)2-1=254,或S7=-2×

2、[1-(-2)7]1-(-2) =-86. 2.已知數(shù)列{an}的前n項(xiàng)和為Sn,滿足a2=15,Sn+1=Sn+3an+6. (1)證明:{an+3}是等比數(shù)列; (2)求數(shù)列{an}的通項(xiàng)公式以及前n項(xiàng)和Sn. (1)證明在Sn+1=Sn+3an+6中,令n=1,得S2=S1+3a1+6, 得a1+a2=a1+3a1+6,即a1+15=4a1+6, 解得a1=3. 因?yàn)镾n+1=Sn+3an+6, 所以an+1=3an+6. 所以an+1+3an+3=3an+9an+3=3. 所以{an+3}是以6為首項(xiàng),3為公比的等比數(shù)列. (2)解由(1)得an+3=6×3n-

3、1=2×3n, 所以an=2×3n-3. ∴Sn=2×(3+32+33+…3n)-3n=2×3×(1-3n)1-3-3n=3n+1-3-3n. 3.設(shè)數(shù)列{an}的前n項(xiàng)和為Sn,Sn=1-an(n∈N*). (1)求數(shù)列{an}的通項(xiàng)公式; (2)設(shè)bn=log2an,求數(shù)列1bnbn+1的前n項(xiàng)和Tn. 解(1)因?yàn)镾n=1-an(n∈N*), 所以Sn-1=1-an-1(n∈N*,且n≥2), 則Sn-Sn-1=(1-an)-(1-an-1)(n∈N*,且n≥2). 即an=12an-1(n∈N*,且n≥2). 因?yàn)镾n=1-an(n∈N*), 所以S1=1-a1=

4、a1,即a1=12. 所以{an}是以12為首項(xiàng),12為公比的等比數(shù)列. 故an=12n(n∈N*). (2)bn=log2an,所以bn=log212n=-n. 所以1bnbn+1=1n(n+1)=1n-1n+1, 故Tn=1-12+12-13+…+1n-1n+1=1-1n+1=nn+1. 4.設(shè)等差數(shù)列{an}的公差為d,d為整數(shù),前n項(xiàng)和為Sn,等比數(shù)列{bn}的公比為q,已知a1=b1,b2=2,d=q,S10=100,n∈N*. (1)求數(shù)列{an}與{bn}的通項(xiàng)公式; (2)設(shè)cn=anbn,求數(shù)列{cn}的前n項(xiàng)和Tn. 解(1)由題意可得10a1+45d=1

5、00,a1d=2, 解得a1=9,d=29(舍去)或a1=1,d=2, 所以an=2n-1,bn=2n-1. (2)∵cn=anbn,cn=2n-12n-1, ∴Tn=1+32+522+723+…+2n-12n-1,① 12Tn=12+322+523+724+925+…+2n-12n,② ①-②可得12Tn=2+12+122+…+12n-2-2n-12n=3-2n+32n,故Tn=6-2n+32n-1. 5.已知正項(xiàng)數(shù)列{an}的前n項(xiàng)和為Sn,滿足2Sn+1=2an2+an(n∈N*). (1)求數(shù)列{an}的通項(xiàng)公式; (2)已知對(duì)于n∈N*,不等式1S1+1S2+1S3

6、+…+1Sn0,所以a1=1, 當(dāng)n≥2時(shí),2Sn+1=2an2+an(n∈N*), 2Sn-1+1=2an-12+an-1(n∈N*), 作差整理,得an+an-1=2(an+an-1)(an-an-1), 因?yàn)閍n>0,故an+an-1>0,所以an-an-1=12, 故數(shù)列{an}為等差數(shù)列,所以an=n+12. (2)由(1)知Sn=n(n+3)4, 所以1Sn=4n(n+3)=431n-1n+3, 從而1S1+1S2+1S3+…+1Sn =431-14+12-15+13-16+…

7、+1n-2-1n+1+1n-1-1n+2+1n-1n+3 =431+12+13-1n+1-1n+2-1n+3=43116-1n+1-1n+2-1n+3<229. 所以M≥229,故M的最小值為229. 6.已知數(shù)列{an}是公比為q的正項(xiàng)等比數(shù)列,{bn}是公差d為負(fù)數(shù)的等差數(shù)列,滿足1a2-1a3=da1,b1+b2+b3=21,b1b2b3=315. (1)求數(shù)列{an}的公比q與數(shù)列{bn}的通項(xiàng)公式; (2)求數(shù)列{|bn|}的前10項(xiàng)和S10. 解(1)由已知,b1+b2+b3=3b2=21,得b2=7, 又b1b2b3=(b2-d)·b2·(b2+d)=(7-d)·7

8、·(7+d)=343-7d2=315, 得d=-2或2(舍),b1=7+2=9,bn=-2n+11. 于是1a2-1a3=-2a1, 又{an}是公比為q的等比數(shù)列,故1a1q-1a1q2=-2a1, 所以,2q2+q-1=0,q=-1(舍)或12. 綜上,q=12,d=-2,bn=11-2n. (2)設(shè){bn}的前n項(xiàng)和為Tn;令bn≥0,11-2n≥0,得n≤5, 于是,S5=T5=5(b1+b5)2=25. 易知,n>6時(shí),bn<0,|b6|+|b7|+…+|b10|=-b6-b7-…-b10=-(b6+b7+…+b10)=-(T10-T5)=-(0-25)=25,所以,

9、S10=50. B組　能力提升 7.已知數(shù)列{an}的前n項(xiàng)和為Sn,點(diǎn)(n,Sn)(n∈N*)在函數(shù)f(x)=12x2+12x的圖象上. (1)求數(shù)列{an}的通項(xiàng)公式; (2)設(shè)數(shù)列1anan+2的前n項(xiàng)和為Tn,不等式Tn>13loga(1-a)對(duì)任意正整數(shù)n恒成立,求實(shí)數(shù)a的取值范圍. 解(1)∵點(diǎn)(n,Sn)在函數(shù)f(x)=12x2+12x的圖象上, ∴Sn=12n2+12n.① 當(dāng)n≥2時(shí),Sn-1=12(n-1)2+12(n-1),② ①-②,得an=n. 當(dāng)n=1時(shí),a1=S1=1,符合上式. ∴an=n(n∈N*). (2)由(1),得1anan+2=1

10、n(n+2)=121n-1n+2, ∴Tn=1a1a3+1a2a4+…+1anan+2=121-13+12-14+…+1n-1n+2=34-121n+1+1n+2. ∵Tn+1-Tn=1(n+1)(n+3)>0, ∴數(shù)列{Tn}單調(diào)遞增, ∴{Tn}中的最小項(xiàng)為T1=13. 要使不等式Tn>13loga(1-a)對(duì)任意正整數(shù)n恒成立,只要13>13loga(1-a), 即loga(1-a)

11、且a1,a2,a3成等差數(shù)列,求λ的值; (2)若{an}的各項(xiàng)均不為零,問(wèn)當(dāng)且僅當(dāng)λ為何值時(shí),a2,a3,…,an,…成等差數(shù)列?試說(shuō)明理由. 解(1)令n=1,2,得λa2=a1+1=2,2λa3=S2+1=a1+a2+1, 又由a1,a2,a3成等差數(shù)列, 所以2a2=a1+a3=1+a3, 解得λ=3±52. (2)當(dāng)且僅當(dāng)λ=12時(shí),a2,a3,…,an,…成等差數(shù)列, 證明如下: 由已知λnan+1=Sn+1,當(dāng)n≥2時(shí),λ(n-1)an=Sn-1+1, 兩式相減得λnan+1-λnan+λan=an, 即λn(an+1-an)=(1-λ)an, 由于{an}

12、的各項(xiàng)均不相等, 所以λn1-λ=anan+1-an(n≥2), 當(dāng)n≥3時(shí),有λ(n-1)1-λ=an-1an-an-1, 兩式相減可得λ1-λ=anan+1-an-an-1an-an-1, ①當(dāng)λ=12,得anan+1-an=an-1an-an-1+1=anan-an-1, 由于an≠0,所以an+1-an=an-an-1, 即2an=an+1+an-1(n≥3), 故a2,a3,…,an,…成等差數(shù)列. ②再證當(dāng)a2,a3,…,an,…成等差數(shù)列時(shí),λ=12, 因?yàn)閍2,a3,…,an,…成等差數(shù)列, 所以an+1-an=an-an-1(n≥3),可得anan+1-an-an-1an-an-1=anan-an-1-an-1an-an-1=1=λ1-λ, 所以λ=12, 所以當(dāng)且僅當(dāng)λ=12時(shí),a2,a3,…,an,…成等差數(shù)列. 8