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2019-2020年高考数学二轮复习限时训练12等差、等比数列及数列求和文1.xx·高考北京卷已知等差数列{an}满足a1+a2=10,a4-a3=
2.1求{an}的通项公式;2设等比数列{bn}满足b2=a3,b3=a7,问b6与{an}的第n项相等?解1∵a4-a3=2,∴d=2,∴a1+a1+d=10,∴a1=4∴an=a1+n-1×d=4+n-1×2=2n+
2.2由1得a3=2×3+2=8,∴b2=8a7=2×7+2=16,b3=16∴公比q==2∴b6=b3·q3=16×23=128∴128=2n+2,∴n=63即b6与a63相等.2.xx·郑州市模拟已知等差数列{an}的各项均为正数,a1=1,且a3,a4+,a11成等比数列.1求{an}的通项公式;2设bn=,求数列{bn}的前n项和Tn.解1设等差数列{an}的公差为d,由题意知d0,因为a3,a4+,a11成等比数列,所以2=a3a11,所以2=1+2d1+10d,即44d2-36d-45=0,所以d=,所以an=.2bn===,所以Tn==.3.xx届石家庄市高中模拟设数列{an}的前n项和为Sn,a1=1,an+1=λSn+1n∈N*,λ≠-1,且a
1、2a
2、a3+3为等差数列{bn}的前三项.1求数列{an},{bn}的通项公式;2求数列{anbn}的前n项和.解1法一∵an+1=λSn+1n∈N*,∴an=λSn-1+1n≥2,∴an+1-an=λan,即an+1=λ+1ana≥2,λ+1≠0,又a1=1,a2=λS1+1=λ+1,∴数列{an}是以1为首项,公比为λ+1的等比数列,∴a3=λ+12,∴4λ+1=1+λ+12+3,整理得λ2-2λ+1=0,解得λ=1,∴an=2n-1,bn=1+3n-1=3n-
2.法二∵a1=1,an+1=λSn+1n∈N*,∴a2=λS1+1=λ+1,a3=λS2+1=λ1+λ+1+1=λ2+2λ+1,∴4λ+1=1+λ2+2λ+1+3,整理得λ2-2λ+1=0,解得λ=1,∴an+1=Sn+1n∈N*,∴an=Sn-1+1n≥2,∴an+1-an=ann≥2,即an+1=2ann≥2,又a1=1,a2=2,∴数列{an}是以1为首项,公比为2的等比数列,∴an=2n-1,bn=1+3n-1=3n-
2.2由1知,anbn=3n-2×2n-1,设Tn为数列{anbn}的前n项和,∴Tn=1×1+4×21+7×22+…+3n-2×2n-1,
①∴2Tn=1×21+4×22+7×23+…+3n-5×2n-1+3n-2×2n.
②①-
②得,-Tn=1×1+3×21+3×22+…+3×2n-1-3n-2×2n=1+3×-3n-2×2n,整理得Tn=3n-5×2n+
5.4.xx·高考湖南卷设数列{an}的前n项和为Sn.已知a1=1,a2=2,且an+2=3Sn-Sn+1+3,n∈N*.1证明an+2=3an;2求Sn.1证明由条件,对任意n∈N*,有an+2=3Sn-Sn+1+3,因而对任意n∈N*,n≥2,有an+1=3Sn-1-Sn+
3.两式相减,得an+2-an+1=3an-an+1,即an+2=3an,n≥
2.又a1=1,a2=2,所以a3=3S1-S2+3=3a1-a1+a2+3=3a
1.故对一切n∈N*,an+2=3an.2解由1知,an≠0,所以=
3.于是数列{a2n-1}是首项a1=1,公比为3的等比数列;数列{a2n}是首项a1=2,公比为3的等比数列.因此a2n-1=3n-1,a2n=2×3n-
1.于是S2n=a1+a2+…+a2n=a1+a3+…+a2n-1+a2+a4+…+a2n=1+3+…+3n-1+21+3+…+3n-1=31+3+…+3n-1=,从而S2n-1=S2n-a2n=-2×3n-1=5×3n-2-1.综上所述,Sn=。