na n(n-1)d 2

an=Sn-S(n-1)=nan-n(n-1)-(n-1)a(n-1)+(n-1)(n-2)an=nan-n(n-1)-(n-1)a(n-1)+(n-1)(n-2)0=nan-an-(n-1)a(n-

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/

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

由题得：Sn=1-nan于是有：S(n-1)=1-(n-1)a(n-1)两式相减得：an=(n-1)a(n-1)-nan移项后有：(n+1)an=(n-1)a(n-1)于是：an=[(n-1)/(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)

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

/>令bn=nan,bn的前n项和为Tn.Tn-T（n-1）=bn=3n（n+1）则an=bn/n=3（n+1）.经检验n=1时也满足.综上,数列{an}的通项公式为an=3（n+1）.其实这个用的就

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

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)

an=(n+1)/n*a(n-1)递推a(n-1)=n/(n-1)*a(n-2)a(n-2)=(n-1)/(n-2)*a(n-3).a2=3/2*a1所有式子乘起来,能约的全约掉,an=(n+1)/2

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)/

Sn-S(n-1)=nan-(n-1)a(n-1)-4n+4=an(n-1)an-(n-1)a(n-1)=4(n-1)an-a(n-1)=4所以an=4n-3

S100+S99=a100-a99-1/2^100-2^99=d-1/2^100-2^99;……S2+S1=a2-a1-1/2^2-1/2=d-1/2^2-1/2;以上全部加起来得:SUM=50d-(