等差数列an的前n项何为sn,已知s3=a2*2

n,an,Sn成等差数列,所以n+Sn=2an,即Sn=2an-n,an+1=Sn+1-Sn=2an+1-n-1-2an+n=2an+1-2an-1化简就是an+1=2an+1an+1+1=2an+2

答案为ASn=((a1+an)/2)*nan=a1+(n-1)d根据上式得出：Sn=(2a1+(n-1)d)*n/2=a1*n+n方*d/2-n*d/2limSn/n方=lim(2a1*n+n方*d-

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

因为Sn-Sn-1=n^2-3n-{(n-1)^2-3(n-1)}=2n-4.又由an=Sn-Sn-1,所以an=2n-4,最后还要验证一下,当n=1时,S1=a1,符合题意.d=an-an-1=2易

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

你数列当中的第五个元素

设第一项为：a1,公差为：d1、S15>0可得到a1>-7d2、a8+a9

（Ⅰ）∵等比数列{an}的前n项和为Sn，S1，S3，S2成等差数列，∴2（a1+a1q+a1q2）=a1+a1+a1q，解得q=-12或q=0（舍）．∴q=-12．（Ⅱ）∵a1-a3=3，q=-12

S10=10(a1+a10)/2=45所以a1+a10=9又a4+a7=a1+a10＝9所以a4+a6=a4+a7-d=9-d所以,少一条件.另：若已知S9=45,则9(a1+a9)/2=45,可得出

由题意可得a1b1=S1T1=524=13，故a1=13b1．设等差数列{an}和{bn}的公差分别为d1 和d2，由S2T2=a1+a1+d 1b1+b1 +d&nbs

1、a4-a1=-9=3dd=-3an=25-3(n-1)=-3n+28an>0-3n+28>0n0,a10S8S9>S10所以n=9.Sn最大2、a2=a1+d=22a20=-60+28=-32有1

看图

唉,你太粗心了吧~我给你修正下(向我现在这样的好人不多了哈哈~!)Sm/Sn=(m^2)/(n^2),求am/an?对吧,很简单的呦am/an＝2am/（2an）＝a1+a2m-1/(a1+a2n-1

S(n)＝n^2－9nS(n-1)=(n-1)^2-9(n-1)=n^2-2n+1-9n+9=n^2-11n+10a(n)=S(n)-S(n-1)=(n^2－9n)-(n^2-11n+10)=2n-1

S1/a1=1S2/a2-S1/a1=(2+d)/(1+d)-1=d/(1+d)S3/a3-S1/a1==(3+3d)/(1+2d)-1=(2+d)/(1+2d)2*d/(1+d)=(2+d)/(1+

Sn=((An+1)/2)^2A1=S1=((A1+1)/2)^2(A1-1)^2=0A1=1Sn=n(A1+An)/2=n(1+An)/2=((An+1)/2)^2(An+1)/2=nAn=2n-1

/>n≥2时,an=Sn/n+2(n-1)Sn=nan-2n(n-1)S(n-1)=(n-1)an-2(n-1)(n-2)Sn-S(n-1)=an=nan-2n(n-1)-(n-1)an+2(n-1)

因为这样求得的d只能保证2a2=a1+a3,也就是前3项成等差数列,并不能保证3项之后.可以以较为普遍的情况来分析.

Sn=n(A1+An)/2设Bn=Sn/n=(A1+An)/2Bn-B(n-1)=(A1+An)/2-[A1+A(n-1)]/2=[An-A(n-1)]/2=d/2=常数∴｛Sn/n}是等差数列

当n=1时,a1=S1=1当n≥2时,an=Sn-S(n-1)=3n²-2n-3(n-1)²+2(n-1)=6n-5∵当n=1时,满足an=6n-5又∵an-a(n-1)=6n-5