a1等于1当n大于等于2时sn的平方

首先先说,该题需要有一个条件就是An和Sn的关系,我姑且猜测是{Sn}为{An}的前n项和.An=（√Sn+√Sn-1）/2Sn-Sn-1=（√Sn+√Sn-1）/2(把Sn看做√Sn的平方)√Sn-

∵n≥2时,an=Sn-Sn-1∴由已知得,Sn2=（Sn-Sn-1）（Sn-1）整理得,Sn-Sn-1=-Sn·Sn-1∴1/Sn-1/Sn-1=（Sn-Sn-1）/（Sn·Sn-1）=-1为常数,

证明：an=（√Sn+√Sn-1）/2=Sn-Sn-1=（√Sn+√Sn-1）(√Sn-√Sn-1）∴√Sn-√Sn-1=1/2(√Sn是等差数列）S1=a1=1,√S1=1,∴√Sn=1+(n-1)

sn-s(n-1)=an=[√sn+√s(n-1)]/2√sn-√s(n-1)=1/2√sn-√s1=(n-1)/2√sn=(n+1)/2√sn为等差数列sn=(n+1)(n+1)/4an=sn-s(

Sn^2=an×(Sn-1/2)=(Sn-Sn-1)×(Sn-1/2)整理,得Sn-1-Sn=2SnSn-1等式两边同除以SnSn-11/Sn-1/Sn-1=2,为定值.1/S1=1/a1=1/1=1

(2)bn=1/(2n-1)(2n+1)=1/2[1/(2n-1)-1/(2n+1)]Tn=1/2{1/1-1/3+1/3-1/5+1/5-1/7+...+1/(2n-1)-1/(2n+1)}=1/2

已知：数列an中a1=1,当n≥2时,其前n项和满足sn²=an[sn-(1/2)]；求：an表达式.代入an=sn-s(n-1)到sn²=an[sn-(1/2)],化简得(1/s

s²n=(sn-s(n-1))(sn-1/2)s²n=s²n-sn/2-s(n-1)*sn+s(n-1)/2sn(1/2+s(n-1))=s(n-1)/2sn=s(n-1

s²n=(sn-s(n-1))(sn-1/2)s²n=s²n-sn/2-s(n-1)*sn+s(n-1)/2sn(1/2+s(n-1))=s(n-1)/2sn=s(n-1

Sn-S(n-1)-2^n=S(n-1)Sn/2^n-S(n-1)/2^(n-1)=1S1=1soSn/2^n=nSn=n*2^nan=Sn-S(n-1)=n*2^n-(n-1)2^(n-1)an/2

an=-Sn.S(n-1)Sn-S(n-1)=-Sn.S(n-1)1/Sn-1/S(n-1)=11/Sn-1/S1=n-11/Sn=nSn=1/n

把an用sn-s（n-1）替代计算

当n大于等于2时,由2Sn=b（Sn-1）+3可化为2（Sn-k）=b（Sn-1-k）其中2k-bk=3,求得k=3/（2-b）所以{Sn-k}是一首项为(S1-k)=（2-b）,公比为b/2的等比数

(1)设Sn-S(n-1)=an原式可化为1/2[Sn-S(n-1)]=-Sn*S(n-1)两边同除-1/2*Sn*S(n-1),可得1/Sn-1/S(n-1)=2设cn=1/Sn,可得当n>或=2时

an=Sn-1Sn-Sn-1=Sn-1Sn=2Sn-1S1=a1=5所以,{Sn}是首项为5,公比为2的等比数列Sn=5*(2^n-1)Sn-1=5*(2^(n-1)-1)an=Sn-Sn-1=5[(

an=sn-s(n-1)代入得Sn=2S(n-1)+2^n,即Sn/2^n=S(n-1)/2^(n-1)+1所以Sn=(n+1/2)*2^n,所以an=Sn-S(n-1)=n*2^n+2^(n-1).

2(S_n)^2=2a_nS_n-a_n=>2S_n(S_n-a_n)=-a_n=>2S_n*S_{n-1}=-a_n2S_n*S_{n-1}=-(S_n-S_{n-1})2=-1/S_{n-1}+1

这个题目是要经过尝试,才能得出结论的.最好自己亲自尝试一下,会比较容易在约分的时候发现规律.因S2=a1+a2,将a1=1/3带入a2=2（S2)^2/(2S2-1)可得a2=-2/15=-2/((2

n>1时sn=an(1-2/sn)=(sn-s(n-1))(1-2/sn)=sn-s(n-1)-2+2s(n-1)/sn整理可得：sn*s(n-1)=2(s(n-1)-sn)1/sn-1/s(n-1)