2019届高考数学二轮复习 高考大题专项练 二 数列（B）理.doc

二数列(B)1.(2018·醴陵模拟)已知正项等比数列{an}中,a1+a2=6,a3+a4=24.(1)求数列{an}的通项公式;(2)数列{bn}满足bn=log2an,求数列{an+bn}.(2018·银川模拟)设{an}是公比不为1的等比数列,其前n项和为Sn,且a5,a3,a4成等差数列.(1)求数列{an}的公比;(2)证明:对任意k∈N*,Sk+2,Sk,Sk+.(2018·益阳模拟)已知{an}是各项均为正数的等差数列,且数列{1anan+1}的前n项和为n2(n+2),n∈N*.(1)求数列{an}的通项公式;(2)若数列{an}的前n项和为Sn,数列{1Sn}的前n项和Tn,求证Tn<.(2018·深圳模拟)已知数列{an}满足a1=1,且an=2an-1+2n(n≥2,且n∈N*),(1)求证:数列{an2n}是等差数列;(2)求数列{an}的通项公式;(3)设数列{an}的前n项之和为Sn,求证:Sn2n>2n-:(1)设数列{an}的首项为a1,公比为q(q>0).则a1+a1·q=6,a1·q2+a1·q3=24,解得a1=2,q=2,所以an=2×2n-1=2n.(2)由(1)得bn=log22n=n,设{an+bn}的前n项和为Sn,则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+.(1)解:设数列{an}的公比为q(q≠0,q≠1),由a5,a3,a4成等差数列,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)证明:法一对任意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,对任意k∈N*,Sk+2,Sk,Sk+∈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,对任意k∈N*,Sk+2,Sk,Sk+.(1)解:由{an}是各项均为正数的等差数列,且数列{1anan+1}的前n项和为n2(n+2),n∈N*,当n=1时,可得1a1a2=12×3=16,①当n=2时,可得1a1a2+1a2a3=22×4=14,②②-①得1a2a3=112,所以a1·(a1+d)=6,③(a1+d)(a1+2d)=12.④由③④解得a1=2,d={an}的通项公式为an=n+1.(2)证明:由(1)