求证数列an减n为等比数列
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Sn=2an+1Sn-1=2a(n-1)+1an=Sn-S(n-1)=2an-2a(n-1)an=2a(n-1)an/a(n-1)=2{an}为等比数列S1=a1=2a1+1a1=-1an=-1*2^
1:令n=1在Sn=2an-2n中有a1=2Sn=2an-2nS(n-1)=2a(n-1)-2(n-1)an=Sn-S(n-1)=2an-2n-(2a(n-1)-2(n-1))=2an-2a(n-1)
Sn=2an-4Sn=2[Sn-S]-4Sn=2S+4Sn+4=2(S+4)所以Sn+4构成公比为2的等比数列Sn+4=(S1+4)*2^(n-1)利用S1=2a1-4=a1求出S1=a1=4Sn+4
题目是这样的吗?已知数列{an}的前n项和为sn,sn=1/3(an-1)(n属于N+)(1)求a1、a2(2)求证数列{an}是等比数列(1):sn=1/3(an-1)n=1s1=a1=1/3(a1
S(n+1)=4An+2(1)S(n)=4A(n-1)+2(n≥2)(2)(1)-(2)得,A(n+1)=4A(n)-4A(n-1)(n≥2)[A(n+1)-2An]/[A(n)-2A(n-1)]=[
令Sn为an前n项和,Sn=n-an,S(n-1)=n-1-a(n-1),两式相减,an=1-an+a(n-1),2(an-1)=a(n-1)-1,所以an-1是公比为1/2的等比数列,a1-1=-1
证明:a1=-1,则a2=-5,所以b1=1,b2=-1.a(n+1)=-an-4n-2bn+1/bn=[a(n+1)+2n]/(an+2n)=(-an-4n-2+2n)/(an+2n)=-1所以{b
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证明:A(n+1)=Sn+3n+1,则An=S(n-1)+3n-2两式想减得A(n+1)-An=Sn+3n+1-(S(n-1)+3n-2)=An+3即A(n+1)+3=2(An+3)即(A(n+1)+
(1)充分性:当t=-1时,Sn=3^n-1,因此a1=S1=2,又当n>=2时,an=Sn-S(n-1)=(3^n-1)-[3^(n-1)-1]=2*3^(n-1),由于a(n+1)/an=(2*3
1.a(n+1)=3an-2则a(n+1)-1=3(an-1)令bn=an-1那么b1=2,b(n+1)/bn=3所以数列{bn}为等比数列,即数列{(an)-1}为等比数列2.b(n+1)=3bn=
方法一:A(n+1)-1=3An-3=3(An-1),且A1-1=2,所以数列{An-1}为公比为3,首项为2的等比数列方法二:设A(n+1)+k=3(an+k),即A(n+1)=3An+2k,则2k
1.证:Sn=(3an-n)/2Sn-1=[3a(n-1)-(n-1)]/2an=Sn-Sn-1=[3an-3a(n-1)-1]/2an=3a(n-1)+1an+1/2=3a(n-1)+3/2=3[a
an+1=4an-3n+1an+1-n-1=4an-4nan+1-(n+1)=4(an-n)[an+1-(n+1)]/(an-n)=4所以an-n是等比数列bn=an-n,所以bn是等比数列,b1=1
an+3^(n+1)=4a(n-1)+3^n+3^(n+1)=4a(n-1)+4*3^n=4(a(n-1)+3^n)所以q=4首项=a1+3=4得证
数列{an}前N项和Sn3Sn=(an-1),(1)当n>=2,有:3Sn-1=[a(n-1)-1],(2)(1)-(2),3an=an-an-1an/an-1=-1/2,(n>=2)当n=1,3S1
2an-2^n=sn2a(n-1)-2^(n-1)=s(n-1)两式想减,有2an-2a(n-1)-2^n+2^(n-1)=an2an-2a(n-1)-2^(n-1)-an=0an-2a(n-1)=2
证明:由已知得:Sn+1=2^nSn=2^n-1an/a(n-1)=[sn-s(n-1]/[s(n-1)-S(n-2)]=[2^n-1-2^(n-1)+1]/[2^(n-1)-1-2^(n-2)+1]
∵Sn=kq^n-k∴S(n+1)=kq^(n+1)-k∴a(n+1)=S(n+1)-Sn=[kq^(n+1)-k]-(kq^n-k)=k[q^(n+1)-q^n]=k[(q-1)q^na(n+1)/
an=1/(-3)^na(n-1)=1/(-3)^(n-1)an/a(n-1)=[1/(-3)^n]/[1/(-3)^(n-1)]=-1/3(常数)是等比数列再问:能讲解一下是用什么来证明的吗?再答: