an sn=4n×5n

当n>1时,an=sn-sn-1代入化简得：1/Sn-1/Sn-1=2所以：1/Sn=2n-1所以：Sn=1/(2n-1)当n>1时,an=sn-sn-1=-2/[(2n-1)(2n-3)]当n=1时

n-1=再问：29/20n-4n=

2Sn²=2anSn-an,知an=Sn-S(n-1)代换后化简可得（过程不难但打起来很闹心……）1/Sn-1/S(n-1)=2故而1/Sn的通项公式为1/Sn=2n-1（1/S1=1）,即

因为S(n+1)-S(n)=A(n+1),根据题意有:2S(n+1)^2=2A(n+1)S(n+1)-A(n+1),将上式代入此式得:2S(n+1)^2=2[S(n+1)-S(n)]S(n+1)-S(

2sn^2=2ansn-an2sn^2=2[sn-s(n-1)]sn-[sn-s(n-1)]2sn^2=2sn^2-2sns(n-1)-sn+s(n-1)2sns(n-1)+sn-s(n-1)=01/

已知：数列{an}中,a1=1,Sn为数列{an}的前n项和,且2an/(anSn-Sn²)=1,(n≥2)；(1).证明数列{1/Sn}是等差数列；(2).求数列{an}的通项公式.(1)

推倒正确,你的问题是?

an=Sn-Sn-1在往下可以得到1/Sn是一个等差数列

太简单了.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

［n/2]+[n/3]+[n/4]+[n/5]+[n/6]=(30n+20n+15n+12n+10n)/60=87n/60=29n/60题目是不是打错了..等于29吧?这样n=60再问：是69~~~└

an=sn-s(n-1),2sn^2=2(sn-sn-1)sn-sn+s(n-1)=2sn^2-2s(n-1)sn-sn+s(n-1)2sns(n-1)=s(n-1)-sn2=1/sn-1/s(n-1

1.证：n≥2时,2Sn²=2anSn-an=2[Sn-S(n-1)]Sn-[Sn-S(n-1)]整理,得S(n-1)-Sn=2SnS(n-1)等式两边同除以SnS(n-1)1/Sn-1/S

2Sn的平方=2anSn-an2Sn(Sn-an)=-an2SnS(n-1)=S(n-1)-Sn1/Sn-1/Sn-1=2｛1/Sn｝等差数列公差2首项1/21/Sn=1/2+2(n-1)=[4n-3

等于呀,你把后面的算式一道前面来n(n+2)(n+4)+1/6)(n-1)n(n+2)(n+4)=n(n+2)(n+4)[1+1/6(n-1)]=1/6n(n+2)(n+4)(n+5)

n=1时,(s1-1)^2=s1*s1即-2s1+1=0解得s1=1/2n=2时,(s2-1)^2=(s2-s1)*s2解得：s2=2/3n=3时,(s3-1)^2=(s3-s2)*s3解得：s3=3

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

由题意知：2an/[anSn-(Sn)²]=1（n>1）则：(Sn)²-anSn+2an=0（n>1）又因为：an=Sn-S(n-1)（n>1）所以：(Sn)²-[Sn-

解4（n+2）=5n4n+8=5nn=8

证明：（1）∵an2-2anSn+1=0，an=Sn-Sn-1（n≥2）∴（Sn-Sn-1）2-2（Sn-Sn-1）Sn+1=0⇒Sn2-Sn-12=1故{Sn2}成等差数列．（2）∵a12-2a12

1/[n(n+1)]=1/n-1/(n+1)套用这个公式1除以（n+3）(n+4)+1除以(n+4)(n+5)+、、、1除以(n+10)(n+11)=1/(n+3)-1/(n+4)+1/(n+4)-1