高二数学(数列的极限)设等差数列{an}的公差d是2,前n项和为sn则lim((an^2-n^2)/sn)=
1个回答
详细解答见图,点击放大,再点击再放大.
图片已经传上,稍等几分钟即可.
相关问题
设等差数列{an}公差d是2,前n项和为Sn,则lim(an^2-n^2)/Sn
等差数列{an}的前n项和为Sn,若lim n→∞ Sn/n^2=2,则该数列的公差d=
设等差数列{an}的公差d是2,前n项的和为Sn,则limn→∞a2n−n2Sn=______.
(2007•天津)设等差数列{an}的公差d是2,前n项的和为Sn,则limn→∞a2n−n2Sn=______.
设sn为等差数列an的前n项和,若a1=1,公差d=2sn+1-sn=36.则n=?
等差数列{an}前n项和为Sn,且Sn=3n^2+n 求公差d
已知数列的前n项和Sn=2^n-1,则lim(an+2)/Sn
在等差数列an中,前n项和为Sn,若Sm=2n,Sn=2m,则公差d的值为
记数列{an}的前n项和为Sn,若{Sn/an}是公差为d的等差数列,则{an}为等差数列时d=
设等差数列{an}的首项a1为a,公差d=2,前n项和为Sn.