设An前n项和Sn且Sn等于2An减2
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![设An前n项和Sn且Sn等于2An减2](/uploads/image/f/7246153-1-3.jpg?t=%E8%AE%BEAn%E5%89%8Dn%E9%A1%B9%E5%92%8CSn%E4%B8%94Sn%E7%AD%89%E4%BA%8E2An%E5%87%8F2)
1.Sn=-2an+3有S(n-1)=-2a(n-1)+3则an=Sn-S(n-1)=-2an+2a(n-1)=>an=a(n-1)*2/3所以,{an}为共比数列,q=2/32.Sn=-2an+3有
(1)由已知有:2a1=4096得a1=2048,又an+sn=4096,an+1+Sn+1=4096,两式相减得an+1=an/2,所以an是以1/2为公比的等比数列,故an=2048*(1/2)^
(本小题满分13分)(I)由题意,当n=1时,得2a1=a1+3,解得a1=3.当n=2时,得2a2=(a1+a2)+5,解得a2=8.当n=3时,得2a3=(a1+a2+a3)+7,解得a3=18.
题目中应该是:S(n+1)=4an+2∵S(n+1)=4an+2Sn=4a(n-1)+2∴S(n+1)-Sn=4an-4a(n-1)a(n+1)=4an-4a(n-1)a(n+1)-2n=2(an-2
S(n+1)+S(n)=2a(n)+1S(n)+S(n-1)=2a(n-1)+1两式相减s(n+1)-s(n-1)=a(n+1)+a(n)=2a(n)-2a(n-1)整理后有a(n+1)-a(n)+2
(1)∵{An}为等比数列,则有An+1=An·q,又∵Sn+1,Sn,Sn+2成等差数列,∴Sn+1+Sn+2=2Sn∴Sn+An+Sn+An+An·q=2Sn∴可得2+q=0所以q=-2(2)这里
2Sn(Sn-An)=-An2SnSn-1=Sn-1-Sn1/Sn-1/Sn-1=2{1/Sn}便是一个等差数列,其首项为1/S1=1/A1=1/2得出的结果便是:Sn=2/(4n-3)An=2/(4
当n=1时、有2s1+1=3a1,即有a1=1,因为2Sn+1=3an,所以2Sn+1+1=3an+1.后式减去前式,得2an+1=3an+1-3an.即有an+1=3an,为等比数列,且公比为3,所
n=an+1S(n+1)=2Sn+n+5.1Sn=2S(n-1)+n-1+5=2S(n-1)+n+4.2(1)-(2)得S(n+1)-Sn=2[Sn-S(n-1)]+1a(n+1)=2an+1a(n+
an=-Sn.S(n-1)Sn-S(n-1)=-Sn.S(n-1)1/Sn-1/S(n-1)=11/Sn-1/S1=n-11/Sn=nSn=1/n
1.sn=2an+ns(n-1)=2a(n-1)+n-1相减得an=2an-2a(n-1)+1整理得an-1=2[2a(n-1)-1]所以an-1是等比数列首项a1由a1=2a1+1得a1=-1所以a
Sn=a1(1-q^n)/(1-q)Sn+1=a1[1-q^(n+1)]/(1-q)Sn+2=a1[1-q^(n+2)]/(1-q)2Sn+2=Sn+Sn+1a1[1-q^(n+1)]/(1-q)+a
因为an,Sn,an^2成等差数列所以2Sn=an^2+an2an=2Sn-2S(n-1)=an^2+an-a(n-1)^2-a(n-1)得:(an-a(n-1))(an+a(n-1))-(an+a(
1)利用Sn+Sn-1=3n²,由归纳法可以得到Sn,其中用到奇数项平方和and偶数项平方和公式,你可以查下2)用an-an-1>0可得a范围再问:其中用到奇数项平方和and偶数项平方和公式
1.a[n]=S[n]-S[n-1]=1/2(√S[n]+√S[n-1])==>√S[n]-√S[n-1]=1/2==>√S[10]-√S[4]=1/2*6=3,√S[4]=√4=2==>√S[10]
a1=1a2=s2-a1=2-1=1a3=s3-a1-a2=4-1-1=2a4=s4-a1-a2-a3=6-1-1-2=2a5=s5-a1-a2-a3-a4=8-1-1-2-2=2a6=s6-a1-a
(1)∵an+Sn=4096,∴a1+S1=4096,a1=2048.当n≥2时,an=Sn-Sn-1=(4096-an)-(4096-an-1)=an-1-an∴anan−1=12an=2048(1
(1)令n=1,得a1=-1.Sn=2an+n,S(n+1)=2a(n+1)+n+1.两式相减,得a(n+1)=2a(n+1)-2an+1.整理得a(n+1)-1=2(an-1),a1-1=-2.综上
解题思路:考查数列的通项,考查等差数列的证明,考查数列的求和,考查存在性问题的探究,考查分离参数法的运用解题过程: