sn求an

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

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

你说的应该是平方和的数列吧.解法如下：a(n)=n^2=n(n+1)-n,n(n+1)=[n(n+1)(n+2)-(n-1)n(n+1)]/3,n=[n(n+1)-(n-1)n]/2,a(n)=[n(

s(n)+s(n+1)=(5/3)a(n+1),s(1)+s(2)=2a(1)+a(2)=(5/3)a(2),2a(1)=(2/3)a(2),a(2)=3a(1)=12.s(n+1)+s(n+2)=(

Sn-S(n-m)=A(n-m+1)+A(n-m+2)+……+A(n-m+m)=b共m项A(n-m+1)=A1+(n-m)dA(n-m+2)=A2+(n-m)d……A(n-m+m)=An=Am+(n-

3乘2的n次方减3.3*2^n-3再问：怎么求、再答：先代入1，因为s1=a1，s1=2a1-3，求出a1等于3，再写一个式子，Sn-1=2a(n-1)-3(n-1)，用第一个式子减这个式子，得到Sn

由题意可得an=2Sn^2/(2Sn-1)又由于an=Sn-S(n-1)即Sn-S(n-1)=2Sn^2/(2Sn-1)化简得Sn+2SnS(n-1)-S(n-1)=0两边同除SnS(n-1)得1/S

n=1时,a1=1+3a1.即a1=-1/2.n>1时,an=Sn-Sn-1=1+3an-（1+3a(n-1））=3an-3a(n-1),即an=3/2a(n-1),即an=-1/2*(3/2)^(n

sn=n^2ans(n-1)=(n-1)^2*a(n-1)sn-s(n-1)=n^2an-(n-1)^2*a(n-1)=an(n^2-1)an=(n-1)^2a(n-1)(n+1)an=(n-1)a(

求出首项a1和公比q代入公式就可以了当q≠1时an=a1q^(n-1)sn=a1(1-q^n)/(1-q)当q=1时an=a1sn=na1

(an+2)/2=√(2Sn)两边平方整理：(an+2)²=8snn-1代换n(a(n-1)+2)²=8s(n-1)两式对应相减(an+2)²-(a(n-1)+2)

已知a_(n+1)=S_n得a_n=S_(n-1)（n＞1）两式相减a_(n+1)-a_n=S_n-S_(n-1)=a_n（n＞1）得a_(n+1)=2a_n（n＞1）因为a_2=S_1=a_1=-2

An=6Sn/(An+3)6Sn=(An)^2+3Ann>=26S(n-1)=(A(n-1))^2+3A(n-1)6An=(An)^2+3An-(A(n-1))^2-3A(n-1)(An)^2-(A(

Sn-a1=48,Sn-an=36,Sn-a1-a2-an-1-an=21,∴2Sn-(a1+an)=84Sn-(a1+an)-(a2+an-1)=21∴2Sn-2Sn/n=84Sn-4Sn/n=21

1）设an=a1*q^(n-1),则有Sn=a1*(1-q^n)/(1-q),[Sn*Sn+2-（Sn+1）^2]=a1^2*{(1-q^n)*[1-q^(n+2)]-[1-q^(n+1)]^2}/(

因为Sn+Sn-1=3an所以Sn-1+Sn-1+an=3an2Sn-1=2anSn-1=an因为Sn=an+1所以Sn-Sn-1=an+1-anan=an+1-an2an=an+1an+1/an=2

Sn-1=(n-1)(n-1)an-1Sn-Sn-1=an=nnan-(n-1)(n-1)an-1(nn-1)an=(n-1)(n-1)an-1an=(n-1)/(n+1)*(n-2)/(n-1)*…

a[n+1]=a[n]/（a[n]+2）是不是这样子?那么两边同时取倒数.1/a[n+1]=[an+2]/an=1+2/an1/a[n+1]+1==2+2/an=2{1/an+1}所以形如1/an+1

an=nsn=n(n+1)/2

已知Sn=2An-1取n=1得:S1=2A1-1又因为S1=A1,解上述方程可得:A1=1Sn=2An-1S(n-1)=2A(n-1)-1注:"n-1"为下标上下两式相减得:Sn-S(n-1)=2An