sn=n² n-1 求它的通项公式

Sn+10=2^(n+1)Sn=-10+2^(n+1)n>=2,则S(n+1)=-10+2^n所以n>=2,an=Sn-S(n-1)=2^(n+1)-2^n=2*2^n-2^n=2^na1=S1=-1

a1=s1=2*1²+1-2=1Sn=2n²+n-2S(n-1)=2(n-1)²+(n-1)-2=2n²-4n+2+n-1-2=2n²-3n-1an=

Sn=3n²+n+1①n=1时S1=a1=3+1+1=5n>=2时S(n-1)=3(n-1)²+(n-1)+1②①-②Sn-S(n-1)=an=3n²+n+1-[3(n-

an=sn-Sn-1(1)Sn=3n^2-nSn-1=3(n-1)^2-(n-1)Sn-Sn-1=3(2n-1)-1=6n-4

当n=1时,A1=S1=2*1-3*1+1=0当n大于等于2时,An=Sn-S（n-1）=(2n^2-3n+1)-[2(n-1)^2-3(n-1)+1]=(2n^2-3n+1)-(2n^2-7n+6)

当n=1时,A1=S1=2*1^2=2；当n>1时：Sn=2*n^2S(n-1)=2*(n-1)^2=2(n^2-2n+1)=2*n^2-4n+2所以An=Sn-S(n-1)=(2*n^2)-(2*n

n>=2S(n-1)=n/(n-1)所以an=Sn-S(n-1)=-1/(n²-n)a1=S1=2/1=2所以an=2,n=1-1/(n²-n),n≥2

a(n)-2^n=(b-1)S(n),ba(1)-2=(b-1)S(1)=(b-1)a(1),a(1)=2.ba(n+1)-2^(n+1)=(b-1)S(n+1),ba(n+1)-2^(n+1)-ba

na(n+1)=S(n)+n(n+1)则(n-1)a(n)=S(n-1)+n(n-1)两式相减得na(n+1)-(n-1)a(n)=a(n)+2nna(n+1)=na(n)+2na(n+1)=a(n)

an=Sn-Sn-1=n(n-1)-(n-1)(n-2)=2n,而a1=2×1=S1=1×(1+1)=2,即n=1时也符合条件;故an=2n

an=Sn-S(n-1)=2n^2-3n-2(n-1)^2+3(n-1)=4n-5an=-1+4(n-1){An}是等差数列,a1=-1,d=4

an=-2[n-(-1/2)^n]=-2n+(-1/2)^(n-1)sn=a1+a2+.+an=-2*1+(-1/2)^(1-1)-2*2+(-1/2)^(2-1)-.-2n+(-1/2)^(n-1)

Sn=3(3^n+1)/2=[3^(n+1)+3]/2S(n-1)=(3^n+3)/2通项公式：an=Sn-S(n-1)=[3^(n+1)+3]/2-(3^n+3)/2=3^n

Sn=nan-n(n-1)an=Sn-S(n-1)=nan-n(n-1)-(n-1)a(n-1)+(n-1)(n-2)化简得(n-1)[an-a(n-1)]=2(n-1)①当n≠1时an-a(n-1)

当n≥2时,an=Sn-S(n-1)=(n²+n-1)-[(n-1)²+(n-1)-1]=2n当n=1时,由Sn=n²+n-1得,a1=S1=1²+1-1=1不

Sn=1-(1/2)bn、S1=b1=1-(1/2)b1,则b1=2/3.b(n+1)=S(n+1)-Sn=(1/2)bn-(1/2)b(n+1),则b(n+1)/bn=1/3.所以,数列{bn}是首

当n≥2an=Sn-S(n-1)=3^n+1-3^(n-1)-1=3*3^(n-1)-3^(n-1)=2*3^(n-1)当n=1a1=S1=4不符合an=2*3^(n-1)所以n=1,an=4n≥2,

2S(n+1)=a(n+1)+3;(1)2S(n)=a(n)+3;(2)(1)-(2)得：2a(n+1)=a(n+1)-a(n);整理得：a(n+1)=-a(n);又:2S(1)=2a(1)=a(1)

Sn=3(2^n+1)/2an=Sn-S(n-1)=3(2^n+1)/2-3[2^(n-1)+1)/2]=[2^n-2(n-1)]*3/2=3*2^(n-2)就是应用了an=Sn-S(n-1)

n≥2则S(n-1)=3^(n-1)+1所以an=Sn-S(n-1)=2*3^(n-1)a1=S1所以a1=3+1=4不符合n≥2时的an2*3^(n-1)所以an=4,n=12*3^(n-1),n≥