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高考大題專項(xiàng)三　高考中的數(shù)列 1.(2018山西呂梁一模,17)已知{an}是首項(xiàng)為1的等比數(shù)列,數(shù)列{bn}滿足b1=2,b2=5,且anbn+1=anbn+an+1. (1)求數(shù)列{an}的通項(xiàng)公式; (2)求數(shù)列{bn}的前n項(xiàng)和. 2.(2018福建龍巖4月質(zhì)檢,17)已知正項(xiàng)等比數(shù)列{an}的前n項(xiàng)和為Sn,且Sn=2an-1(n∈N+). (1)求數(shù)列{an}的通項(xiàng)公式; (2)若bn=lg an,求數(shù)列{an+bn}的前n項(xiàng)和Tn. 3.(2018北京海淀期末,15)已知等差數(shù)列{an}的前n項(xiàng)和Sn,且a2=5,S3=a7. (1)數(shù)列{an}的通項(xiàng)公式; (2)若bn=2an,求數(shù)列{an+bn}的前n項(xiàng)和. 4.(2018河北唐山一模,17)已知數(shù)列{an}為單調(diào)遞增數(shù)列,Sn為其前n項(xiàng)和,2Sn=an2+n. (1)求{an}的通項(xiàng)公式; (2)若bn=an+22n+1anan+1,Tn為數(shù)列{bn}的前n項(xiàng)和,證明:Tn<. 5.(2018湖南衡陽二模,17)等差數(shù)列{an}中,a3=1,a7=9,Sn為等比數(shù)列{bn}的前n項(xiàng)和,且b1=2,若4S1,3S2,2S3成等差數(shù)列. (1)求數(shù)列{an},{bn}的通項(xiàng)公式; (2)設(shè)cn=|an|bn,求數(shù)列{cn}的前n項(xiàng)和Tn. 6.已知數(shù)列{an}的前n項(xiàng)和為Sn,Sn=(m+1)-man對任意的n∈N+都成立,其中m為常數(shù),且m<-1. (1)求證:數(shù)列{an}是等比數(shù)列; (2)記數(shù)列{an}的公比為q,設(shè)q=f(m),若數(shù)列{bn}滿足b1=a1,bn=f(bn-1)(n≥2,n∈N+).求證:數(shù)列1bn是等差數(shù)列; (3)在(2)的條件下,設(shè)cn=bnbn+1,數(shù)列{cn}的前n項(xiàng)和為Tn,求證:Tn<1. 7.(2018宿州十三所中學(xué)期中,17)已知數(shù)列{an}的前n項(xiàng)和為Sn,并且滿足a1=1,nan+1=Sn+n(n+1). (1)求數(shù)列{an}的通項(xiàng)公式; (2)若bn=an2n,數(shù)列{bn}的前n項(xiàng)和為Tn,求Tn; (3)在(2)的條件下,是否存在常數(shù)λ,使得數(shù)列Tn+λan+2為等比數(shù)列?若存在,試求出λ;若不存在,說明理由.  高考大題專項(xiàng)三　高考中的數(shù)列 1.解 (1)把n=1代入已知等式得a1b2=a1b1+a2,∴a2=a1b2-a1b1=3a1. ∴{an}是首項(xiàng)為1,公比為3的等比數(shù)列,即an=3n-1. (2)由已知得bn+1-bn=an+1an=3, ∴{bn}是首項(xiàng)為2,公差為3的等差數(shù)列,其通項(xiàng)公式為bn=3n-1, ∴Sn=n(b1+bn)2=n(2+3n-1)2=3n2+n2. 2.解 (1)由Sn=2an-1(n∈N+),可得S1=2a1-1, ∴a1=2a1-1,∴a1=1. 又S2=2a2-1,∴a1+a2=2a2-1,∴a2=2. ∵數(shù)列{an}是等比數(shù)列, ∴公比q=a2a1=2, ∴數(shù)列{an}的通項(xiàng)公式為an=2n-1. (2)由(1)知,bn=lg an=(n-1)lg 2, ∴Tn=(b1+a1)+(b2+a2)+…+(bn+an) =(0+1)+(lg 2+2)+…+[(n-1)lg 2+2n-1] =[lg 2+2lg 2+…+(n-1)lg 2]+(1+2+…+2n-1) =n(n-1)2lg 2+2n-1. 3.解 (1)設(shè)等差數(shù)列{an}的首項(xiàng)為a1,公差為d, 則a1+d=5,3a1+3d=a1+6d,解得a1=3,d=2. 由an=a1+(n-1)d,則an=2n+1. 因此,通項(xiàng)公式為an=2n+1. (2)由(1)可知:an=2n+1, 則bn=22n+1,bn+1bn=22(n+1)+122n+1=4. 因?yàn)閎1=23=8,所以{bn}是首項(xiàng)為8,公比為q=4的等比數(shù)列. 記{an+bn}的前n項(xiàng)和為Tn, 則Tn=(a1+b1)+(a2+b2)+…+(an+bn)=(a1+a2+…+an)+(b1+b2+…+bn)=n(a1+an)2+b1(1-qn)1-q=n2+2n+8(4n-1)3. 4.(1)解 當(dāng)n=1時,2S1=2a1=a12+1, 所以(a1-1)2=0,即a1=1. 又{an}為單調(diào)遞增數(shù)列,所以an≥1. 由2Sn=an2+n得2Sn+1=an+12+n+1, 所以2Sn+1-2Sn=an+12-an2+1, 整理得2an+1=an+12-an2+1, 即an2=(an+1-1)2, 所以an=an+1-1,即an+1-an=1, 所以{an}是以1為首項(xiàng),1為公差的等差數(shù)列,所以an=n. (2)證明 因?yàn)閎n=an+22n+1anan+1=n+22n+1n(n+1)=12nn-12n+1(n+1), 所以Tn=1211-1222+1222-1233+…+12nn-12n+1(n+1)=1211-12n+1(n+1)<12. 5.解 (1)在等差數(shù)列{an}中,設(shè)公差為d,則a7-a3=4d=9-1=8,故d=2, ∴an=a3+(n-3)d=1+2(n-3)=2n-5. 設(shè)等比數(shù)列{bn}的公比為q,依題意有6S2=4S1+2S3,故q=2,∴bn=2n. (2)∵cn=|2n-5|2n. 當(dāng)n=1時,T1=6, 當(dāng)n=2時,T2=10, 當(dāng)n≥3時,2n-5>0, Tn=10+123+324+…+(2n-7)2n-1+(2n-5)2n, ① 2Tn=20+124+325+…+(2n-7)2n+(2n-5)2n+1, ② ①-②,得-Tn=-10+8+2(24+…+2n)-(2n-5)2n+1, ∴Tn=34+(2n-7)2n+1. ∴Tn=6,n=1,10,n=2,34+(2n-7)2n+1,n≥3. 6.證明 (1)當(dāng)n=1時,a1=S1=1. ∵Sn=(m+1)-man, ① ∴Sn-1=(m+1)-man-1(n≥2), ② 由①-②,得an=man-1-man(n≥2), 即(m+1)an=man-1. ∵a1≠0,m<-1,∴an-1≠0,m+1≠0. ∴anan-1=mm+1(n≥2). ∴數(shù)列{an}是首項(xiàng)為1,公比為mm+1的等比數(shù)列. (2)∵f(m)=mm+1,b1=a1=1,bn=f(bn-1)=bn-1bn-1+1(n≥2), ∴1bn=bn-1+1bn-1(n≥2), ∴1bn-1bn-1=1(n≥2), ∴數(shù)列1bn是首項(xiàng)為1,公差為1的等差數(shù)列. (3)由(2)得1bn=n,則bn=1n,故cn=bnbn+1=1n(n+1), 因此,Tn=112+123+…+1n(n+1)=11-12+12-13+13-14+…+1n-1n+1=1-1n+1<1. 7.解 (1)∵nan+1=Sn+n(n+1), ① ∴當(dāng)n≥2時,(n-1)an=Sn-1+n(n-1), ② 由①-②可得an+1-an=2(n≥2),且a1=1,a2=S1+1(1+1)=3, ∴數(shù)列{an}是首項(xiàng)為1,公差為2的等差數(shù)列,即an=2n-1. (2)由(1)知數(shù)列an=2n-1,∴bn=2n-12n, 則Tn=121+322+523+…+2n-32n-1+2n-12n, ① ∴12Tn=122+323+524+…+2n-32n+2n-12n+1, ② 由①-②得, 12Tn=12+2122+123+…+12n-2n-12n+1 =12+214(1-12n-1)1-12-2n-12n+1, ∴Tn=3-2n+32n. (3)由(2)知Tn=3-2n+32n, ∴Tn+λan+2=(3-2n+32n+λ)2n+3=(3+λ)2n+3-12n, ∴要使數(shù)列Tn+λan+2為等比數(shù)列, 當(dāng)且僅當(dāng)3+λ=0,即λ=-3. 故存在λ=-3,使得數(shù)列Tn+λan+2為等比數(shù)列.