1、素养提升3高考中数列解答题的提分策略1.2020山西大学附属中学校诊断,12分已知等比数列an的前n项和为Sn(nN*),-2S2,S3,4S4成等差数列,且a2+2a3+a4 =116.(1)求数列an的通项公式;(2)若bn =-(n+2)log2|an|,求数列1bn的前n项和Tn.2.原创题,12分已知正项数列an的前n项和为Sn,a1 =1,Sn2 =an+12-Sn+1,其中为常数.(1)证明:Sn+1 =2Sn+.(2)是否存在实数,使得数列an为等比数列?若存在,求出;若不存在,请说明理由.3.12分已知数列an的前n项和为Sn,a1 =1,且满足Sn =an+1.(1)求数列
2、an的通项公式;(2)求数列nan的前n项和Tn.4.12分已知数列an的各项均为正数,且an2-2nan-(2n+1) =0,nN*.(1)求数列an的通项公式;(2)若bn =(-1)n-1an,求数列bn的前n项和Tn.素养提升3高考中数列解答题的提分策略1.(1)设等比数列an的公比为q,由 - 2S2,S3,4S4成等差数列,可知q1,2S3=4S4 - 2S2,即2a1(1 - q3)1 - q=4a1(1 - q4)1 - q - 2a1(1 - q2)1 - q,化简得2q2 - q - 1=0,解得q= - 12,a2+2a3+a4=116,即 - 12a1+214a1 -
3、18a1=116,解得a1= - 12,则an=( - 12)n,nN*.(5分)(2)bn= - (n+2)log2|an|= - (n+2)log212n=n(n+2),可得1bn=1n(n+2)=12(1n - 1n+2),则Tn=12(1 - 13+12 - 14+1n - 1 - 1n+1+1n - 1n+2)=12(1+12 - 1n+1 - 1n+2)=34 - 12(1n+1+1n+2).(12分)2.(1)an+1=Sn+1 - Sn,Sn2=an+12 - Sn+1,Sn2=(Sn+1 - Sn)2 - Sn+1,(1分)Sn+1(Sn+1 - 2Sn - )=0.(3分)
4、an0,Sn+10,Sn+1 - 2Sn - =0,Sn+1=2Sn+.(5分)(2)Sn+1=2Sn+,Sn=2Sn - 1+(n2),两式相减,得an+1=2an(n2).(8分)S2=2S1+,即a2+a1=2a1+,a2=1+,由a20,得 - 1.若an是等比数列,则a1a3=a22,(10分)即2(+1)=(+1)2,得=1.(11分)经检验,=1符合题意.故存在=1,使得数列an为等比数列.(12分)3.(1)Sn=an+1,当n=1时,a2=1,当n2时,Sn - 1=an,an=Sn - Sn - 1=an+1 - an(n2),an+1=2an(n2),a1=1,a2=1,
5、不满足上式,数列an是从第二项起的等比数列,公比为2,an=1,n=1,2n - 2,n2.(6分)(2)由(1)知,当n=1时,T1=1,当n2时,Tn=1+220+321+n2n - 2,2Tn=12+221+322+n2n - 1, - Tn=1+21+22+2n - 2 - n2n - 1=1 - 2n - 11 - 2 - n2n - 1,Tn=(n - 1)2n - 1+1.当n=1时也满足上式,综上,Tn=(n - 1)2n - 1+1.(12分)4.(1)由an2 - 2nan - (2n+1)=0得an - (2n+1)(an+1)=0,所以an=2n+1或an= - 1.又
6、数列an的各项均为正数,所以an=2n+1,nN*.(5分)(2)由(1)知an=2n+1,nN*,bn=( - 1)n - 1an=( - 1)n - 1(2n+1),所以Tn=3 - 5+7 - 9+( - 1)n - 1(2n+1),故 - Tn= - 3+5 - 7+9 - +( - 1)n - 1(2n - 1)+( - 1)n(2n+1), - 得,2Tn=3 - 21 - 1+1 - 1+( - 1)n - 2 - ( - 1)n(2n+1)=3 - 211 - ( - 1)n - 11 - ( - 1) - ( - 1)n(2n+1)=2+( - 1)n - 1 - ( - 1)n(2n+1)=2+( - 1)n - 1(2n+2),所以Tn=1+( - 1)n - 1(n+1).(12分)(12分)
