等差数列{an} 前n项和为sn,则lim(n趋向无穷)【sn/(an^2)】=?
来源:学生作业帮 编辑:作业帮 分类:综合作业 时间:2024/08/27 01:11:40
等差数列{an} 前n项和为sn,则lim(n趋向无穷)【sn/(an^2)】=?
a(n)=a+(n-1)d,[a(n)]^2=[a+(n-1)d]^2=[nd+a-d]^2,
s(n)=na+n(n-1)d/2=(d/2)n^2+n(a-d/2),
当d=0时,a(n)=a,s(n)=na.
当d=0,a=0时,s(n)/[a(n)]^2没有意义.
当d=0,a不等于0时,s(n)/[a(n)]^2=n/a,lim(n->无穷大){s(n)/[a(n)]^2}=(n->无穷大){n/a}=正无穷大.
d不等于0时,
s(n)/[a(n)]^2=[(d/2)n^2+n(a-d/2)]/[nd+a-d]^2=[d/2+(a-d/2)/n]/[d+(a-d)/n]^2,
lim(n->无穷大){s(n)/[a(n)]^2}=lim(n->无穷大){[d/2+(a-d/2)/n]/[d+(a-d)/n]^2}=(d/2)/d^2=1/(2d)
s(n)=na+n(n-1)d/2=(d/2)n^2+n(a-d/2),
当d=0时,a(n)=a,s(n)=na.
当d=0,a=0时,s(n)/[a(n)]^2没有意义.
当d=0,a不等于0时,s(n)/[a(n)]^2=n/a,lim(n->无穷大){s(n)/[a(n)]^2}=(n->无穷大){n/a}=正无穷大.
d不等于0时,
s(n)/[a(n)]^2=[(d/2)n^2+n(a-d/2)]/[nd+a-d]^2=[d/2+(a-d/2)/n]/[d+(a-d)/n]^2,
lim(n->无穷大){s(n)/[a(n)]^2}=lim(n->无穷大){[d/2+(a-d/2)/n]/[d+(a-d)/n]^2}=(d/2)/d^2=1/(2d)
等差数列{An}的前n项和为Sn,若 lim Sn/n方 =2
设等差数列an的公差d为2,前n项和为Sn,则lim(an^2-n^2)/Sn=?
高中数学有关数列、极限的题:设等差数列{an}的公差d=2,前n项的和为sn,则lim(an^-n^0/sn=
正项无穷等比数列{an}前n项和为Sn,lim(Sn/Sn+1)=1 求公比范围
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
已知等差数列{an}中,a1=2,d=-2,前n项的和为 Sn,则 Sn( )
已知数列{an}是首项为1,公差为2的等差数列,Sn(n属于N)是数列的前n项和,则lim下面为n到无穷 Sn/n^2减
已知等差数列{an}的前n项和为Sn,且a1不等于0,求(n*an)/Sn的极限、(Sn+Sn+1)/(Sn+Sn-1)
等差数列an,bn的前n项和分别为Sn,若Sn/Tn=2n/(3n+1),求lim an/bn
设Sn为等差数列{an}的前n项和,若a1=1,公差d=2,Sn+2-Sn=36,则n=( )
已知等差数列{an}的前N项和为Sn,a1=-2/3,满足Sn+1/Sn+2=an(n大于等于2)
已知等差数列{an}的前N项和为Sn,a1=-2/3,满足Sn+1/Sn+2=an(n大于等于2),