an方 an=4Sn 3

an=a(n-1)+1/(n²-n)=a(n-1)+1/(n-1)-1/nan+1/n=a(n-1)+1/(n-1)an+1/n=a(n-1)+1/(n-1)=a(n-2)+1/(n-2)=

a(n+1)/an=5^nan=a1*(a2/a1)(a3/a2)(a4/a3).(an/an-1)=4*5¹5²5³.*5^(n-1)=4*5^[1+2+3+……(n-

一个白痴

解题思路：利用an=Sn-Sn-1来解答。解题过程：最终答案：略

an+1项应该是平方吧如果是的话,解如下：分解因式：（an+1+an）（（n+1）an+1-nan）=0an+1=-an或者an+1=nan/(n+1)(1)当an+1=-an的,an=（-1）^(n

a(n+1)=3an+4.1a(n+2)=3a(n+1)+4.22-1a(n+2)=4a(n+1)-3an由特征方程得x^2=4x-3x=1或3an=A1^n+B3^na1=1,a2=7A=-2,B=

1.bn=(3an-2)/(an-1)an=(bn-2)/(bn-3)a(n+1)=[b(n+1)-2]/[b(n+1)-3]a(n+1)=(4an-2)/(3an-1)3a(n+1)an-a(n+1

sn=2n^2-n+2s(n-1)=2(n-1)^2-(n-1)+2两式相减an=4n-3

n=1时,a1=S1=4×1²+2×1=6n≥2时,an=Sn-S(n-1)=4n²+2n-[4(n-1)²+2(n-1)]=8n-2n=1时,a1=8×1-2=6,同样

令f(x)=(x+4)/(2x-1)=x,解得：x1=-1,x2=2取F(x)=(x+1)/(x-2)则：F^-1(x)=(2x+1)/(x-1),那么g(x)=F.f.F^-1=(x+1)/(x-2

a(n+1)-an=b(n+1)/2的n+1次方=2n次方是对2吧,也就是说分母是2的n次方,对吧!如果对2,那么bn=2的n+1次方（n>1）,b1=2,Sn=（2的n+2次方）-6

a(n+2)+2an=3a(n+1)a(n+2)-a(n+1)=2a(n+1)-2an[a(n+2)-a(n+1)]/[a(n+1)-2an]=2∴数列{an+1-an}是等比数列a(n+1)-an=

a(n+1)=4an+9(n+1)表示下标a(n+1)+3=4(an+3)[a(n+1)+3]/(an+3)=4所以数列{an+3}是以a1+3=5为首相q=4为公比的等比数列an+3=5*(4)^(

你把An^2看成是Bn嘛,那么{Bn}就是一个公差为4的等差数列,求出Bn再开平方就行了

令bn=an²则b(n+1)=bn+4所以bn是等差数列,d=4b1=a1²=1所以bn=4n-3an>0所以an=√(4n-3)

1.a(n+1)^2=an^2+4,令bn=an^2,b(n+1)=bn+4,b1=a1^2=1bn是一个等差数列,其通项bn=4(n-1)+1=4n-3因an>0,an=√(4n-3)2.在数列{a

Sn=n^2+2n-1,S1=1^2+2-1=2an=Sn-S(n-1)=n^2+2n-1-[(n-1)^2+2(n-1)-1]=2n+1a1=3所以a1≠S1当n=1时an=2当n>1时an=2n+

a100=S100-S99=4*100^2-100-4*99^2+99=4*199-1=795

a1=1a2=(3+2)/(1+4)=1……an=1则bn=2

an=Sn-S(n-1)=2^n-1-[2^(n-1)-1]=2^(n-1)an^2=4^(n-1)a1^2=1a1^2+a2^2+...+an^2=(1-4^n)/(1-4)=(4^n-1)/3