(-1)nan-an-1=2n(n>2)S100=

A(n+1)表示第n+1项(n+1)A(n+1)^2-nAn^2+A(n+1)An=0n(A(n+1)+An)(A(n+1)-An)+A(n+1)(A(n+1)+An)=0(A(n+1)+An)[nA

a1+2a2+3a3+...+nan=n(n+2)(n+1)a1+2a2+3a3+...+(n-1)a(n-1)=(n-1)n(n+1)nan-(n-1)a(n-1)=3n(n+1)nan=(n-1)

nA(n+1)=(n+2)An+n可变形为n[A(n+1)+(n+1)]=(n+2)[An+n]∴[A(n+1)+(n+1)]/[An+n]=(n+2)/n构造数列{Tn},使Tn=An+n则T1=A

两边除以n(n+1)a(n+1)/(n+1)=-an/n+1两边减去1/2a(n+1)/(n+1)-1/2=-an/n+1/2a(n+1)/(n+1)-1/2=-(an/n-1/2)所以an/n-1/

以下用a(n)表示数列的第n项.题目中的式子是a(n)=a(n-1)/{[(-1)^n]×a(n-1)-2}的意思吧?（1）由a(n)=a(n-1)/[(-1)^n×a(n-1)-2],两边取倒数,得

liman＝lim[(2n+1)an]/(2n+1)＝lim[(2n+1)an]×lim1/(2n+1)＝3×0＝0所以,3＝lim[(2n+1)an]＝2×limnan＋liman＝2×limnan

na(n+1)=(n+1)an+2n[a(n+1)+t]=(n+1)(an+t)t=2[a(n+1)+2]/(an+2)=(n+1)/n(a2+2)/(a1+2)=2/1(a3+2)/(a2+2)=3

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

第一问,根据前几个,猜想通项公式为n,验证.第二问,注意到n(n+1)/2为n的前n项和.考察an-nan-n=（（n+1）an-1²-(n-1)an-1+n）/(an-1²+1)

a1+2a2+3a3+...+nan=n(n+1)(n+2).①a1+2a2+3a3+...+(n-1)an-1=(n-1)n(n+1).②①-②得：nan=n(n+1)(n+2)-(n-1)n(n+

a1+2a2+3a3+...+(n-1)a(n-1)=(n-1)n(n+1)a1+2a2+3a3+...+nan=n(n+1)(n+2)2试-1式得nan=3n(n+1)an=3(n+1)

a1+2a2+3a3+...+nan=n(n+1)*(n+2),则：a1+2a2+3a3+...+（n-1）×an-1=n(n-1)*(n+1),两式相减：nan=n(n+1)*(n+2)-n(n-1

令n=1时,a1=1*2*3=6；依题意：a1+2a2+3a3+.+nan=n(n+1)(n+2),a1+2a2+3a3+.+nan+(n+1)a(n+1)=(n+1)(n+2)(n+3)两式相减,得

由题意可知,An=-2/n,故极限(3n+1)An=(3n+1)*(-2/n)=(-6n-2)/n=-6-2/n=-6,这里极限就是N趋向于无穷大时,而2/n当n趋向于无穷大时的值为零

用累积法做,由A(n+1)/An=(n+2)/n得A(n)/A（n-1)=(n+1)/n-1A(n-1)/A(n-2)=n/n-2A(n-2)/A(n-3)=n-1/n-3A(n-3)/A(n-4)=

∵数列{a[n]}满足a[1]+2a[2]+3a[3]+...+na[n]=(n+1)(n+2)∴a[1]+2a[2]+3a[3]+...+na[n]+(n+1)a[n+1]=(n+2)(n+3)将上

a1+2a2+3a3+~+nan=n(n+1)(2n+1)知,a1+2a2+3a3+~+(n-1)an-1=(n-1)n(2n-1),n≠1时两式相减知an=(n+1)(2n+1)-(n-1)(2n-

(1)设{nan}数列的前n项和为Sn,则Sn=a1+2a2+3a3+.+nan=n(2n+1)=2n^2+n所以S(n-1)=(n-1)[2(n-1)+1]=2n^2-3n+1所以nan=Sn-S(

已知数列{an}满足a1=1,a(n+1)=nan(1)求{an}的通项公式；(2)证明:1/a1+1/a2+.+1/an≤3-(1/2)^(n-2).(1)因为a(n+1)=nan,即a(n+1)/

再问：你好厉害再答：谢谢夸奖再问：我还没讲题目呢再答：题目是什么？不是求an吗？再问：再问：再问：如果知道的话，明天早上6点一定要发给我哦，谢谢！不知道的话就算了，我要睡觉了！拜拜再答：你一题一题拍，