求证数列1 Sn是等车数列
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由Sn^2=an(Sn-1/2),两边同时除以Sn,拆开括号,得到Sn=an-an/2Sn,移项,an-Sn=an/2Sn,两边同时除以an,乘以2,得到2(an-Sn)/an=1/Sn,那么1/(S
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
an+2Sn*Sn-1=0其中an=Sn-Sn-1代入上式:Sn-Sn-1+2Sn*Sn-1=0a1=1/2,故Sn和Sn-1≠0,上式两边同除以Sn*Sn-1得:1/Sn-1-1/Sn+2=0即:1
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
(1)∵a(n+1)=2an+1∴a[n+1]+1=2a[n]+2=2(a[n]+1)∴a[n]+1为等比数列,等比=2(2)a[n]+1=(a[1]+1)*2^(n-1)=2^n∴a[n]=-1+2
由Sn=Sn-1/2Sn-1+1,两边同时取倒数可得1/Sn=(2Sn-1+1)/Sn-11/Sn=2+1/Sn-1即1/Sn-1/Sn-1=2故{1/Sn}是首项为1/2,公差为2的等差数列1/Sn
Sn=n-anS(n-1)=(n-1)-a(n-1)两式作差得:an=1+a(n-1)-an整理得:2(an-1)=a(n-1)-1即2bn=b(n-1)再问:已知数列{an}的前n项和为Sn,且对任
第一个搞定我就不罗嗦了即1/Sn-1/Sn-1=2所以有1/Sn-1/Sn-1=21/Sn-1-1/Sn-2=21/Sn-2-1/Sn-3=2…………1/S2-1/S1=2叠加得1/Sn-1/S1=2
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
∵点(an,sn)在直线y=1/2(x2+x)上∴Sn=1/2(an^2+an)∴an=Sn-S(n-1)=1/2(an^2+an)-1/2(a[n-1]^2+a[n-1])即1/2(an^2-an)
∵a(n+1)=(n+2)Sn/n且a(n+1)=S(n+1)-Sn∴S(n+1)-Sn=(n+2)*Sn/n∴S(n+1)=[(n+2)/n+1]Sn=(2n+2)/n*Sn∴S(n+1)/(n+1
a(n+1)=(a-1)Sn+2an=(a-1)S(n-1)+2两式想减,得a(n+1)-an=(a-1)an即a(n+1)=a*an,以上的讨论中n的范围是n>1且n∈N,∵a>1∴a(n+1)/a
log2(S1+1)=log2(a1+1)=1a1=1log2(Sn+1)=log2(2^n),Sn+1=2^n,Sn=2^n-1a(n)=Sn-S(n-1)=2^n-2^(n-1)-1+1=2^(n
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)+
简单的要死,你成绩在学校排中等吗?log2(Sn+1)=n,所以Sn+1=2^n,Sn=2^n-1,an=Sn-S(n-1)=(2^n-1)-(2^(n-1)-1)=2^(n-1)a(n+1)/an=
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
且S1=2,S<n1>-Sn=Sn2=bn这句话的意思没看明白!∵bn=Sn+2∴b(n+1)=S(n+1)+2b(n+1)-bn=S(n+1)-Sn=bn∴b(n+1)=2*bn则b(n+1)/bn
n>1时sn=an(1-2/sn)=(sn-s(n-1))(1-2/sn)=sn-s(n-1)-2+2s(n-1)/sn整理可得:sn*s(n-1)=2(s(n-1)-sn)1/sn-1/s(n-1)
Sn=3(an-1)/2S(n+1)=3[a(n+1)-1]/2两式相减,得a(n+1)=3[a(n+1)-an]/2a(n+1)=3*an,n∈N+∴a(n+1)/an=3=常数∴数列{an}是等比
Sn=1/8*(an+2)^2Sn+1=1/8*(an+1+2)^2Sn+1-Sn=1/8*[(an+1^2+4an+1)-(an^2+4an)]8an+1=(an+1^2+4an+1)-(an^2+