已知数列an前n项和是sn,且2sn 3=3an 求数列an的通项公试

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已知数列{an}前n项和为Sn,且Sn=-2an+3

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

已知数列an是等差数列,且a1不等于0,Sn为这个数列的前n项和,求limnan/Sn.limSn+Sn-1/Sn+Sn

1、Sn=(a1+an)n/2所以nan/Sn=2an/(a1+an)=2[a1+(n-1)d]/[2a1+(n-1)d]上下除以(n-1)=2[a1/(n-1)+d]/[2a1/(n-1)+d]n-

已知数列an是等差数列,且a1≠0,Sn为这个数列的前n项和.求1、lim nan/Sn 2、lim (Sn+Sn+1)

1、Sn=(a1+an)n/2所以nan/Sn=2an/(a1+an)=2[a1+(n-1)d]/[2a1+(n-1)d]上下除以(n-1)=2[a1/(n-1)+d]/[2a1/(n-1)+d]n-

已知数列{an}的前n项和为Sn,且Sn=n-5an-85,n属于正整数

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

已知数列{an}的前n项和为Sn,且满足an+2Sn*Sn-1=0,a1=1/2.求证:{1/Sn}是等差数列

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

已知数列{an}的前n项和为Sn,且满足Sn=Sn-1/2Sn-1 +1,a1=2,求证{1/Sn}是等差数列

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

已知数列an前n项和是sn,且2sn+3=3an 求数列an的通项公试

2Sn+3=3An2Sn=3An-312S(n-1)=3A(n-1)-321式-2式得2[Sn-S(n-1)]=3An-3A(n-1)2An=3An-3A(n-1)An=3A(n-1)An/A(n-1

已知数列{an}的前n项和为Sn,且Sn=n-5an-85,n∈N*

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

已知数列{an}的前n项和为Sn,且Sn=n-5an-85,n∈N*

(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{

已知Sn是数列{an}的前n项和,且有Sn=n2+1,则数列{an}的通项an=______.

a1=S1=1+1=2,an=Sn-Sn-1=(n2+1)-[(n-1)2+1]=2n-1,当n=1时,2n-1=1≠a1,∴an=2,n=12n−1,n≥2.答案:an=2,n=12n−1,n≥2.

已知数列(an),Sn是前n项的和,且an=S(n-1)+2,a1=2

(1)∵a[n]=S[n-1]+2(n>=2)∴a[n+1]=S[n]+2将上述两式相减,得:a[n+1]-a[n]=a[n]即:a[n+1]/a[n]=2∵a[1]=2∴a[n]是首项和公比都是2的

已知数列{an}的前n项和是Sn,且Sn=2an-n(n∈N*).

(1)∵Sn=2an-n,∴n=1时,a1=2a1-1,解得a1=1.n≥2时,an=Sn-Sn-1=(2an-n)-(2an-1-n+1)=2an-2an-1-1,∴an=2an-1+1,∴an+1

已知数列{An},Sn是其前n项和,且满足3An=2Sn+n,n为正整数,求证数列{An+1/2}为等比数列

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

已知数列an的前n项和为sn,且sn+an=n^2+3n+5/2,证明数列{an-n}是等比数列

Sn+an=n^2+3n+5/2①当n=1时,S1+a1=1^2+3*1+5/2=13/2而S1=a1,所以2a1=13/2,即a1=13/4,所以a1-1=9/4;又S(n-1)+a(n-1)=(n

已知数列{an}中,a2=2,前n项和为Sn,且Sn=n(an+1)/2证明数列{an+1-an}是等差数列

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(

已知数列中各项均为正数,sn是数列an 中的前N项和,且Sn=1/2.求数列an的通项公式

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

已知数列{an}的前n项和为Sn,且Sn=23an+1(n∈N*);

(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,∴

已知数列{an}的各项均为正数,Sn是数列{an}的前n项和,且4Sn=an2+2an-3.

(1)当n=1时,a1=s1=14a21+12a1−34,解出a1=3,又4Sn=an2+2an-3①当n≥2时4sn-1=an-12+2an-1-3②①-②4an=an2-an-12+2(an-an

已知数列{an}的前n项和是Sn(n∈N^*),a1=1且Sn*SN-1+1/2an=0.

n≥2时,SnS(n-1)+(1/2)an=02SnS(n-1)+Sn-S(n-1)=0等式两边同除以SnS(n-1)2+1/S(n-1)-1/Sn=01/Sn-1/S(n-1)=2,为定值1/S1=

已知数列an其前n项和为Sn,且Sn=3n^2+5n,求证数列an是等差数列

因为Sn=3n^2+5nS(n-1)=3(n-1)^2+5(n-1)两式相减所以an=6n-3+5=6n+2所以an=8+6(n-1),所以an是以8为第一项,公差为6的等差数列.