Sn为前n项和,且满足S2n-1=1 2an

因为数列{an}为等差数列,且a1=1,则由等差数列性质可得:前n项和Sn=a1n-(n(n-1)/2)*D即Sn=n-(n(n-1)/2)*D,S2n=2n-(2n(2n-1)/2)*D且S2n/S

an=a1q^(n-1)Sn=a1(q^n-1)/(q-1)(Sn)^2+(S(2n))^2=[a1(q^n-1)/(q-1)]^2+[a1(q^(2n)-1)/(q-1)]^2=[a1/(q-1)]

S2n表示的是前2n项的和,不是原Sn的每2项为一组的新序列的和；（1-1/n)^-n.

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

(1)、S2n-1=1/2an^2和an是各项均不为0的等差数列得S1=1/2a1^2=a1a1=2S3=1/2a2^2=3a2a2=6所以an=4n-2n为偶数时bn=1/2an-1=2n-3(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

(1):因为数列{an}为等差数列,且a1=1,则由等差数列性质可得:前n项和Sn=a1n-(n(n-1)/2)*D即Sn=n-(n(n-1)/2)*D,S2n=2n-(2n(2n-1)/2)*D且S

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+

S2n/Sn=(4n+2)/(n+1),即S2/S1=6/2S2=3S1=3a1=3a2=S2-a1=3-2=1d=2-1=1an=n

由Sn=n-Sa知,an=Sn-Sn-1=1(>=2).a1=1-Sa

取n=1,则s2-2*s1=1,即：a1+a2-2a1=1,又an为等差数列,故d=1,所以an=n.看看是不是这样.

取n=1,则s2-2*s1=1,即：a1+a2-2a1=1,a2=2又an为等差数列,故d=a2-a1=1,所以an=n

竟然第一问会做,那么an=n就不给你说了bn=np^(n)那么Tn=p+2p^2+3p^3+.np^n①则pTn=p^2+2p^3+.(n-1)p^n+np^(n+1)②由①-②得(1-p)Tn=p+

当n=1时,有S2/S1=4因为s1=1所以S2=4=a1+a2=1+a2所以a2=3d=2又当假设成立时,sn=n*a1+(n-1)*n=n*nS2n=4*n*nS2N/SN=4成立.所以An=a1

1.若首项为a1的等差数列{an}为奥运数列,试求出数列{an}的通项公式.S1/S2=a1/(a1+a2)=(a1+a2)/(a1+...+a4)=S2/S4(a1+a2)^2=a1(a1+...+

设公差为d根据Sn=n(a1+an)/2＝na1+n(n-1)d/2＝n[2+(n-1)d]/2S2n/Sn＝2[2+(2n-1)d]/[2+(n-1)d]与原式比较,即：2[2+(2n-1)d]/[

s1=a1=1n=1时,S2＝s1*(4+2)/(1+1)=3a2=s2-s1=2公差d=a2-a1=1故an=n

S2n=2n+n*(2n-1)dSn=n+n(n-1)d/24Sn=4n+2(n^2-n)dS2n/Sn=4S2n=4Sn4n+2d(n^2-n)=2n+(2n^2-n)d整理,得dn=2nd=2S2

(1)S(2n)=2n(a1+a2n)/2=n(a1+a2n)Sn=n(a1+an)/2S(2n)/Sn=2(a1+a2n)/(a1+an)=4a1+a2n=2a1+2ana2n=an+nd,其中d为