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二　數(shù)列(B) 1.(2018醴陵模擬)已知正項(xiàng)等比數(shù)列{an}中,a1+a2=6,a3+a4=24. (1)求數(shù)列{an}的通項(xiàng)公式; (2)數(shù)列{bn}滿足bn=log2an,求數(shù)列{an+bn}的前n項(xiàng)和Tn. 2.(2018銀川模擬)設(shè){an}是公比不為1的等比數(shù)列,其前n項(xiàng)和為Sn,且a5,a3,a4成等差數(shù)列. (1)求數(shù)列{an}的公比; (2)證明:對(duì)任意k∈N*,Sk+2,Sk,Sk+1成等差數(shù)列. 3.(2018益陽(yáng)模擬)已知{an}是各項(xiàng)均為正數(shù)的等差數(shù)列,且數(shù)列{1anan+1}的前n項(xiàng)和為n2(n+2),n∈N*. (1)求數(shù)列{an}的通項(xiàng)公式; (2)若數(shù)列{an}的前n項(xiàng)和為Sn,數(shù)列{1Sn}的前n項(xiàng)和Tn,求證Tn<119. 4.(2018深圳模擬)已知數(shù)列{an}滿足a1=1,且an=2an-1+2n(n≥2,且n∈N*), (1)求證:數(shù)列{an2n}是等差數(shù)列; (2)求數(shù)列{an}的通項(xiàng)公式; (3)設(shè)數(shù)列{an}的前n項(xiàng)之和為Sn,求證:Sn2n>2n-3. 1.解:(1)設(shè)數(shù)列{an}的首項(xiàng)為a1,公比為q(q>0). 則a1+a1q=6,a1q2+a1q3=24, 解得a1=2,q=2, 所以an=22n-1=2n. (2)由(1)得bn=log22n=n, 設(shè){an+bn}的前n項(xiàng)和為Sn, 則Sn=(a1+b1)+(a2+b2)+…+(an+bn) =(a1+a2+…+an)+(b1+b2+…+bn) =(2+22+…+2n)+(1+2+…+n) =2(2n-1)2-1+n(1+n)2 =2n+1-2+12n2+12n. 2.(1)解:設(shè)數(shù)列{an}的公比為q(q≠0,q≠1), 由a5,a3,a4成等差數(shù)列,得2a3=a5+a4, 即2a1q2=a1q4+a1q3, 由a1≠0,q≠0,得q2+q-2=0, 解得q1=-2,q2=1(舍去),所以q=-2. (2)證明:法一　對(duì)任意k∈N*, Sk+2+Sk+1-2Sk=(Sk+2-Sk)+(Sk+1-Sk) =ak+1+ak+2+ak+1 =2ak+1+ak+1(-2) =0, 所以,對(duì)任意k∈N*,Sk+2,Sk,Sk+1成等差數(shù)列. 法二　對(duì)任意k∈N*, 2Sk=2a1(1-qk)1-q, Sk+2+Sk+1=a1(1-qk+2)1-q+a1(1-qk+1)1-q =a1(2-qk+2-qk+1)1-q, 2Sk-(Sk+2+Sk+1)=2a1(1-qk)1-q-a1(2-qk+2-qk+1)1-q =a11-q[2(1-qk)-(2-qk+2-qk+1)] =a1qk1-q(q2+q-2) =0, 因此,對(duì)任意k∈N*,Sk+2,Sk,Sk+1成等差數(shù)列. 3.(1)解:由{an}是各項(xiàng)均為正數(shù)的等差數(shù)列,且數(shù)列{1anan+1}的前n項(xiàng)和為n2(n+2),n∈N*, 當(dāng)n=1時(shí),可得1a1a2=123=16, ① 當(dāng)n=2時(shí),可得1a1a2+1a2a3=224=14, ② ②-①得1a2a3=112, 所以a1(a1+d)=6, ③ (a1+d)(a1+2d)=12. ④ 由③④解得a1=2,d=1. 所以數(shù)列{an}的通項(xiàng)公式為an=n+1. (2)證明:由(1)可得Sn=n(n+3)2, 那么1Sn=2n(n+3)=23(1n-1n+3). 所以數(shù)列{1Sn}的前n項(xiàng)和Tn=23(1-14+12-15+13-16+14-17+…+1n-1n+3) =23(1+12+13-1n+1-1n+2-1n+3) =23(116-1n+1-1n+2-1n+3) =119-23(1n+1+1n+2+1n+3),n∈N*, 所以Tn<119. 4.(1)證明:因?yàn)閍n=2an-1+2n(n≥2,且n∈N*), 所以an2n=an-12n-1+1,即an2n-an-12n-1=1(n≥2,且n∈N*), 所以數(shù)列{an2n}是等差數(shù)列,公差d=1,首項(xiàng)為a12=12. (2)解:由(1)得an2n=12+(n-1)1=n-12, 所以an=(n-12)2n. (3)證明:因?yàn)镾n=1221+3222+5223+…+(n-12)2n, ① 所以2Sn=1222+3223+5224+…+(n-12)2n+1, ② ①-②得 -Sn=1+22+23+…+2n-(n-12)2n+1=2+22+23+…+2n-(n-12)2n+1-1=2(1-2n)1-2-(n-12)2n+1-1=(3-2n)2n-3. Sn=(2n-3)2n+3,則Sn2n=(2n-3)+32n>2n-3, 所以Sn2n>2n-3.