An=1 更号n 更号n 1,求Sn

用错位相减法a1=1*2^0a2=2*2^1a3=3*2^2.an=n*2^(n-1)Sn=1*2^0+2*2^1+3*2^2+.+n*2^(n-1)2Sn=1*2^1+2*2^2+3*2^3+.+(

当n=1时、有2s1+1=3a1,即有a1=1,因为2Sn+1=3an,所以2Sn+1+1=3an+1.后式减去前式,得2an+1=3an+1-3an.即有an+1=3an,为等比数列,且公比为3,所

1）形如1+1/2+1/3+…+1/n+…的级数称为调和级数(还可以推广到等差数列的倒数之和)；也是P-级数(自然数数列的整数p次幂的倒数之和)的特例；黎曼zeta函数也由此得来.（2）Euler(欧

由题得：an=1/2(1/n-1/(n+1);所以：a1=1/2(1-1/2);a2=1/2(1/2-1/3);a3=1/2(1/3-1/4);.an=1/2(1/n-1/(n+1);sn=a1+a2

n=1时,a1=1+3a1.即a1=-1/2.n>1时,an=Sn-Sn-1=1+3an-（1+3a(n-1））=3an-3a(n-1),即an=3/2a(n-1),即an=-1/2*(3/2)^(n

an=（2^n-1)n=2^n*n-n,令Tn=2^1*1+2^*2+…2^n*n,①则2Tn=2^2*1+2^3*2+…+2^n*(n-1)+2^(n+1)*n②②-①得Tn=-(2^1+2^2+…

an=(2n+1)*3^na1=3*3^1a2=5*3^2a3=7*3^3.an=(2n+1)*3^nSn=3*3^1+5*3^2+7*3^3+.(2n+1)*3^n3Sn=3*3^2+5*3^3+7

n=1时,S1+a1=2a1=(1-1)/(1×2)=0a1=0n≥2时,Sn+an=(n-1)/[n(n+1)]=n/[n(n+1)]-1/[n(n+1)]=1/(n+1)-[1/n-1/(n+1)

因为Sn=n+(1/2)an,所以Sn+1=n+1+(1/2)an+1,两式相减得an+1=1+(1/2)an+1-(1/2)an,所以整理后可得an+1=-an+2,当n=1时,a1=2,所以a2=

an=1/n(n+1)(n+2)=[1/n(n+1)-1/(n+1)(n+2)]/2,a1=1/6所以S1=a1=1/6n>=2时,Sn=a1+a2+...+an=[1/1*2-1/2*3]/2+[1

已知a_(n+1)=S_n得a_n=S_(n-1)（n＞1）两式相减a_(n+1)-a_n=S_n-S_(n-1)=a_n（n＞1）得a_(n+1)=2a_n（n＞1）因为a_2=S_1=a_1=-2

sn=a1+a2+a3+……+an=1*2^0+2*2+3*2^2+4*2^3+……+n2^(n-1)2sn=1*2+2*2^2+3*2^3+……+n*2^n两式相减得-sn=1+2+2^2+2^3+

由题意,S(n)-S(n-1)=2a(n+1)-2a(n),即a(n)=2a(n+1)-2a(n),于是a(n+1)=a(n)*3/2,即a(n)是公比是q=3/2的等比数列,且首项是a(1)=1,所

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)

Sn-1=(n-1)(n-1)an-1Sn-Sn-1=an=nnan-(n-1)(n-1)an-1(nn-1)an=(n-1)(n-1)an-1an=(n-1)/(n+1)*(n-2)/(n-1)*…

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).

a[n+1]=a[n]/（a[n]+2）是不是这样子?那么两边同时取倒数.1/a[n+1]=[an+2]/an=1+2/an1/a[n+1]+1==2+2/an=2{1/an+1}所以形如1/an+1

数列{A(n)},A1=1,A(n+1)=3A(n)+4.求A(n)和S(n).1.A(n+1)=3A(n)+4--->A(n)=3A(n-1)+4==3[3A(n-2)+4]+4==(3^2)A(n

解题思路：裂项相消法解题过程：an=1/n(n+2)=1/2n-1/2(n+2)sn=1/2-1/2*3+1/4-1/2*4+1/2*3-1/2*5..........+1/2(n-2)-1/2(n)

已知Sn=2An-1取n=1得:S1=2A1-1又因为S1=A1,解上述方程可得:A1=1Sn=2An-1S(n-1)=2A(n-1)-1注:"n-1"为下标上下两式相减得:Sn-S(n-1)=2An