一直数列an的前n项和s n等于n方减3n-搜答案-神马作文网

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

a4=S4-S3=64-27=37

第一题,n=10时,Sn=-(a1+a2+a3+……）+2（a1+a2+……+a9)=-(9+10-n)n/2+90=(n^2-19n)/2+90.第二题实在是看不清楚你是怎么样写的题目第三题：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

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

由sn=2n/n+1得sn-1=2(n-1)/nan=sn-sn-1∴an=2/[n（n+1)]a1=s1=1满足an=2/[n（n+1)]∴1/a8=36

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Sn=5+9+……+(4n+1)=n(5+4n+1)/2=n(4n+6)/2=2n²+3n

∵an＝1n(n+1)=1n−1n+1，∴S5=a1+a2+a3+a4+a5=1−12+12−13+13−14+14−15+15−16＝56，故选B

（1）由Sn^2=an(Sn-1/2),an=Sn-Sn-1(n≥2)得Sn^2=(Sn-Sn-1)(Sn-1/2)即2Sn-1Sn=Sn-1-Sn.由题意知Sn-1Sn≠0,上式两边同除以Sn-1S

A1=S1=1/3(A1-1)3A1=A1-1A1=-1/2S2=A1+A2=1/3(A2-1)-3/2+3A2=A2-1A2=-1/4再问：求证数列an为等比数列再答：Sn-S(n-1)=an所以1

an=Sn-Sn-1=-SnS(n-1)(Sn-Sn-1)/[SnS(n-1)]=-11/S(n-1)-1/Sn=-11/Sn-1/S(n-1)=1,为定值.1/S1=1/a1=1/(1/2)=2数列

An+2Sn*Sn-1=0Sn-Sn-1+2Sn*Sn-1=01/Sn-1-1/Sn+2=01/Sn=2nSn=1/2n(n>=2)An=1/(2n-2n^2)(n>=2)=1/2(n=1)

1.Sn=n-5an-85Sn-1=n-1-5a(n-1)-85an=Sn-Sn-1=1-5an+5a(n-1)则6an=5a(n-1)+1∴6an-6=5a(n-1)-5即(an-1)/[a(n-1

显然可递推求出:因为sn+1/sn=an-2=sn-s(n-1)-2,所以有1/sn=-s（n-1)-2,进而有sn=1/[-s(n-1)-2],据s1=a1=-1/2,得出:s2=-2/3,进而反复

解题思路：方法：数列通项的求法：已知sn,求an。求和：错位相减法。解题过程：

a1=1a2=s2-a1=2-1=1a3=s3-a1-a2=4-1-1=2a4=s4-a1-a2-a3=6-1-1-2=2a5=s5-a1-a2-a3-a4=8-1-1-2-2=2a6=s6-a1-a

an=n(2^n-1)an=n*2^n-na1=1*2^1-1a2=2*2^2-2a3=3*3^3-3.an=n*2^n-nSn=a1+a2+a3+.+an=1*2^1-1+2*2^2-2+3*3^3