某等差数列,lim（n接近无限大）Sn/（an）²=1/6,求公差.

来源：学生作业帮助网编辑：作业帮时间：2024/05/06 16:51:35

某等差数列,lim（n接近无限大）Sn/（an）²=1/6,求公差.
某等差数列,lim（n接近无限大）Sn/（an）²=1/6,求公差.

某等差数列,lim（n接近无限大）Sn/（an）²=1/6,求公差.
设公差为d,则
an=a1+(n-1)d=dn+a1-d
Sn=na1+n(n-1)d/2
∴Sn/a²n=[na1+n(n-1)d/2]/(dn+a1-d)²
∴lim（n-->∞）Sn/（an）²
=lim（n-->∞）[na1+n(n-1)d/2]/(dn+a1-d)²
=lim（n-->∞）[na1+n(n-1)d/2]/(d²n²+2d(a1-d)n+(a1-d)²]
=lim（n-->∞）[a1/n+(1-1/n)d/2]/(d²+2d(a1-d)/n+(a1-d)²/n²]
=(d/2)/d²=1/(2d)
∵lim（n接近无限大）Sn/（an）²=1/6
∴1/(2d)=1/6 ∴d=3

你把，登差的设出来

Sn=(a1+an)*n/2=na1+n(n-1)d/2
an=a1+(n-1)d
Sn/(an)^2=(na1+n(n-1)d/2)/(a1+(n-1)d)^2
=(na1+n^2d/2-nd/2)/(a1^2+2(n-1)a1d+(n-1)^2d^2) 上下除以n^2
=(d/2+a1/n-d/2n)/((1-1/n)^2*d^2+2(n-1)a1d/n^2+a1^2/n^2)
=(d/2)/(d^2) (n→无穷)
=1/2d
所以d=3

某等差数列,lim（n接近无限大）Sn/（an）²=1/6,求公差. 某等差数列,lim（n接近无限大）Sn/（an）²=1/6,求公差d. 数列 1/（4n²-1）的前n项的和,Sn（n接近无限大）等於? 等差数列{an} 前n项和为sn,则lim（n趋向无穷）【sn/（an^2）】=? 等差数列{an}前n项和Sn 已知lim [Sn/(n²+1）]=-a1/8（a1>0) 则Sn达到最大值时的n=__ 等差数列{an}前n项和Sn 已知lim [Sn/(n²+1）]=-a1/8（a1>0) 则Sn达到最大值时的n=__ 设等差数列an前n项和为Sn,若a6=S3=12,则lim（Sn/n平方）= 等差数列{an}的前n项和为Sn,公差为2,则lim（n到无穷大）Sn/ an^2= 设等差数列{an}公差d=2,前n项和Sn,则lim（an^2-n^2）/Sn=? 高二无穷数列极限{an}是等差数列,Sn为数列前n项和(a1≠0)求:（1） lim n→∞ (nan)/Sn（2）求lim n→∞ Sn+Sn+1/Sn+Sn-1(n+1和n-1是角标）第1题我做好了,是2, lim(n→无限大）An=1+1/3lim(n→无限大）*(1+4/9+(4/9)^2...+(4/9)^(n-2)) 设等差数列{an}公差d是2,前n项和为Sn,则lim(an^2-n^2)/Sn 设等差数列{an}公差d是2,前n项和为Sn,则lim(an^2-n^2)/Sn 数列an是等差数列,首项不为0,求lim[Sn+S(n+1)]/[Sn-S(n+1)] 已知数列an中,an=1/(n(n+1)×(n+2)),则lim(n趋向于无限大)Sn的值 {an}是等差数列,a1不等于0,Sn是它前n项的和.求lim (n→∞) (Sn+S(n+1))/(Sn+S(n-1)) 数列an是等差数列,首项不为0,求lim(n*an/Sn) an是等差数列,求lim (Sn+Sn+1)/(Sn+Sn-1)lim (Sn+Sn+1)/(Sn+Sn-1)=[n(n+1)/2+(n+1)(n+2)/2]/[n(n+1)/2+n(n-1)/2]=(2n²+4n+2)/2n²=1+2/n+1/n²我就想知道第一步怎么来的