正项数列an中前n项和为Sn且a1=2,an=2根号2Sn-1 2(n≥2)
来源:学生作业帮助网 编辑:作业帮 时间:2024/09/21 03:27:36
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
因为An+1=2SnAn=2S(n-1)所以A(n+1)-An=2AnA(n+1)/An=3是公比为3,首项a1=1的等比数列,An=A1*q^(n-1)即An=3^(n-1)
an=2[根号(2Sn-1)]+2(n≥2)则an=2sqrt(2S(n-1))+2① ∵Sn=S(n-1)+an=S(n-1)+2sqrt(2S(n-1))+2 =(sqrt(S(
由题意得,Sn=[(an+1)/2]^2①则S(n+1)=[(a(n+1)+1)/2]^2②②-①得(结合a(n+1)=S(n+1)-Sn)a(n+1)=[(a(n+1)+1)/2]^2-[(an+1
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
由题意得,Sn=[(an+1)/2]^2①则S(n+1)=[(a(n+1)+1)/2]^2②②-①得(结合a(n+1)=S(n+1)-Sn)a(n+1)=[(a(n+1)+1)/2]^2-[(an+1
S[1]=a[1]=1/2(a[1]+1/a[1]),于是:a[1]=1=√1-√0S[2]=a[2]+1=1/2(a[2]+1/a[2]),于是:a[2]=√2-1,S[2]=√2S[3]=a[3]
再答:求好评,给一个好评吧。再问:谢谢你啦再答:给好评呀。再问:太棒了再答:不是这个,是按那个问题已解决。再答:谢谢。再答:知道为什么我用了X么?
Sn=n-5an-85(1)S(n+1)=n+1-5a(n+1)-85(2)(2)-(1)整理得6a(n+1)=1+5an即a(n+1)-1=(5/6)(an-1)又由S1=a1=1-5a1-85得a
因为:An+1=2Sn,则A(n-1)+1=2S(n-1)那么:2Sn-2S(n-1)=(An+1)-(A(n-1)+1)(n>=2)又因为:2Sn-2S(n-1)=2An(n>=2)所以:2An=(
∵2根号Sn=an+14Sn=(an+1)^2①4S(n-1)=[a(n-1)]^2②①-②,可得:4an=[an^2-a(n-1)^2]+2[an-a(n-1)]化简可得:2[a(n-1)+an]=
(1)证明:∵Sn=n-5an-85,n∈N*(1)∴Sn+1=(n+1)-5an+1-85(2),由(2)-(1)可得:an+1=1-5(an+1-an),即:an+1-1=56(an-1),从而{
(1)当n=1时,a1=S1=13(a1−1),得a1=−12;当n=2时,S2=a1+a2=13(a2−1),得a2=14,同理可得a3=−18.(2)当n≥2时,an=Sn−Sn−1=13(an−
1、an=Sn-S(n-1)所以2Sn-S(n-1)=20482Sn=S(n-1)+20482Sn-4096=S(n-1)+2048-40962(Sn-2048)=S(n-1)-2048(Sn-204
(一)(1)由a1=1,S(n+1)=4an+2.可得:a1=1,a2=5,a3=16.a4=44.∴由bn=(an)/2^n得:b1=2/4,b2=5/4,b3=8/4,b4=11/4.显然,b1,
Sn=n(an+1)/2S(n+1)=(n+1)[a(n+1)+1]/2用下式减上式a(n+1)=[(n+1)a(n+1)-nan+1]/2即2a(n+1)=[(n+1)a(n+1)-nan+1]即(
当n=1时,S1=a1=1/2(a1^2+a1),解得a1=1当n>1时,an=Sn-S(n-1)=1/2(an^2+an)-1/2[a(n-1)^2+a(n-1)],整理得[an+a(n-1)][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(
(An)^2=2Sn-An=>(A(n-1))^2=2S(n-1)-A(n-1)=>(An)^2-(A(n-1))^2=2Sn-An-2S(n-1)+A(n-1)=>(An+A(n-1))*(An-A
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: