sn是数列的前n项和 且满足sn=3n² 8n-4 则an=?
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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-
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-
1)an=n2)(2)n次方是什么意思?2的n次方吗?再问:是的,麻烦请给详细过程再答:1)An=Sn-S(n-1)=[n(n+1)]/2-[(n-1)n]/2=n2)Tn=2+2*(2)2+3*(2
2·a(n)=2[Sn-S(n-1)]=(n+1)an-n·a(n-1)∴(n-1)an=n·a(n-1),∴an/[a(n-1)]=n/(n-1),.,a3/a2=3/2,a2/a1=2/1,将上述
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
S1=A1=2A1-3故A1=3而An=Sn-S(n-1)=(2An-3n)-[2A(n-1)-3(n-1)]=2An-2A(n-1)-3故An=2A(n-1)+3故An+3=2[A(n-1)+3]即
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
证明1/Sn等差数列,即证1/Sn-1/Sn-1=常数.当n=2时,a2+2(a1+a2)*a1=0,a2=-1/4,所以1/S2-1/S1=2.n>=2时,an=Sn-Sn-1带入an+2Sn*Sn
(1)∵数列a[n]的前n项和为S[n],且满足a[n]+2S[n]S[n-1]=0,n≥2∴S[n]-S[n-1]+2S[n]S[n-1]=0两边除以S[n]S[n-1],得:1/S[n-1]-1/
由题得:Sn=1-nan于是有:S(n-1)=1-(n-1)a(n-1)两式相减得:an=(n-1)a(n-1)-nan移项后有:(n+1)an=(n-1)a(n-1)于是:an=[(n-1)/(n+
(1)∵Sn-Sn-1=2SnSn-1∴1Sn−1−1Sn=2即1Sn−1Sn−1=−2(常数)∴{1Sn}为等差数列  
解题思路:其他............................................................解题过程:同学你好,能否把题目写清楚一点
由Sn=n-Sa知,an=Sn-Sn-1=1(>=2).a1=1-Sa
应该是a1=0.5吧.(1)先把a1转化,Sn-(Sn-1)+2Sn*Sn-1=0,(Sn-1)-Sn=2Sn*Sn-1因为Sn不为0,所以两边同除Sn*Sn-1可得1/Sn-1/(Sn-1)=2很明
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
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
An=Sn-S(n-1)=n^2+n-(n-1)^2-n+1=n^2+n-n^2+2n-1-n+1=2nA(n+1)=2(n+1)bn=1/4(1/n-1/(n+1)Tn=1/4(1/1-1/2+1/
(1)6a1=a1^2+3a1+2解得a1=1或2(2)6sn=an^2+3an+26s(n-1)=a(n-1)^2+3a(n-1)+2两式想减得6an=an^2-a(n-1)^2+3an-3a(n-
(1)3an=2Sn+n...①3an+1=2Sn+1+n+1...②②-①得:3an+1-3an=2an+1+1即an+1=3an+1==>an+1+1/2=3(an+1/2)an+1+1/2/an