已知{an}前n项和为Sn,且满足Sn-Sn-1 2SnSn-1=0()
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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
a(1)=s(1)=1-5a(1)-85,6a(1)=-84,a(1)=-14.a(n+1)=s(n+1)-s(n)=(n+1)-5a(n+1)-85-[n-5a(n)-85]=1-5a(n+1)+5
Sn+1/(2n+1)-Sn/(2n-1)=1Sn/(2n-1)=S1+n-1→Sn=(S1+n-1)(2n-1)→Sn=n(2n-1)an=4n-31/√an=2/2√(4n-3)>2/(√4n-3
设:等差数列{an}的公差为d,通项为an=a1+(n-1)d,则:sn=a1+a2+...+an=na1+n(n-1)d/2lim(n->∞)(n*an)/Sn=lim(n->∞)[n*(a1+(n
1.a(n+1)=sn/2,a(n+2)=s(n+1)/2,后式减前式得:a(n+2)-a(n+1)=a(n+1)/2,a(n+2)/a(n+1)=3/2,数列a(n+1)为公比q=3/2,首项a2=
∵a(n+1)=1/2Sn.∴n≥2时,an=1/2S(n-1)∴a(n+1)-an=1/2[Sn-S(n-1)]=1/2an∴a(n+1)=3/2an∴a(n+1)/an=3/2∵a1=1,∴a2=
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
由Sn=n-Sa知,an=Sn-Sn-1=1(>=2).a1=1-Sa
1.n=1时,a1=S1=1²+1=2n≥2时,Sn=n²+nS(n-1)=(n-1)²+(n-1)an=Sn-S(n-1)=n²+n-(n-1)²-
A1=S1=1/3(A1-1)3A1=A1-1A1=-1/2S2=A1+A2=1/3(A2-1)-3/2+3A2=A2-1A2=-1/4再问:求证数列an为等比数列再答:Sn-S(n-1)=an所以1
Sn=n-5an-85则an=Sn-S(n-1)=n-5an-85-(n-1)+5a(n-1)+85=1-5an+5a(n-1)即6an=5a(n-1)+16an-6=5a(n-1)+1-66(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{
Sn^2-n^2×Sn-(n^2+1)=0(Sn+1)[Sn-(n^2+1)]=0数列各项为非零实数,S1≠0,且Sn不恒为0,因此只有Sn=n^2+1n=1时,a1=S1=1+1=2n≥2时,an=
(Ⅰ)a1=3,当n≥2时,Sn−1=23an−1+1,∴n≥2时,an=Sn−Sn−1=23an−23an−1,∴n≥2时,anan−1=−2∴数列an是首项为a1=3,公比为q=-2的等比数列,∴
2Sn=n²+n则n≥2时2S(n-1)=(n-1)²+(n-1)=n²-n相减2an=2nan=n2a1=2S1=1+1=2a1=1符合n≥2的式子所以an=n
n=b1.q^(n-1)bn=an-3nan=bn+3n=b1.q^(n-1)+3nSn=a1+a2+...+an=b1(q^n-1)/(q-1)+3n(n+1)/2
S[n]=n-5a[n]-85其中:为了表示清楚,[n]表示下标,S[n-1]=n-1-5a[n-1]-85两式相减:a[n]=1+5(a[n-1]-a[n])a[n]-1=5(a[n-1]-1)-5
a(n+1)=1/3Snsn=3a(n+1)s(n-1)=3anan=sn-s(n-1)=3a(n+1)-3ana(n+1)/an=4/3an为首相1公比4/3等比a1,a3,a5,.a2n-1为首相
an+Sn=41a(n+1)+S(n+1)=2a(n+1)+Sn=422-1得2a(n+1)-an=0a(n+1)=1/2anan+Sn=4an≠0a(n+1)/an=1/2数列{an}是等比数列