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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的前n项和Tn.2.(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kN*,Sk+2,Sk,Sk+1成等差数列.3.(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),nN*.(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Tn2n-3.1.解:(1)设数列an的首项为a1,公比为q(q0).则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,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+12n.2.(1)解:设数列an的公比为q(q0,q1),由a5,a3,a4成等差数列,得2a3=a5+a4,即2a1q2=a1q4+a1q3,由a10,q0,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kN*,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kN*,Sk+2,Sk,Sk+1成等差数列.法二对任意kN*,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-q2(1-qk)-(2-qk+2-qk+1)=a1qk1-q(q2+q-2)=0,因此,对任意kN*,Sk+2,Sk,Sk+1成等差数列.3.(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),nN*,当n=1时,可得1a1a2=123=16, 当n=2时,可得1a1a2+1a2a3=224=14, -得1a2a3=112,所以a1(a1+d)=6, (a1+d)(a1+2d)=12. 由解得a1=2,d=1.所以数列an的通项公式为an=n+1.(2)证明:由(1)可得Sn=n(n+3)2,那么1Sn=2n(n+3)=23(1n-1n+3).所以数列1Sn的前n项和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),nN*,所以Tn2n-3,所以Sn2n2n-3.
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