问题
填空题
计算
|
答案
∵1+2+3+…+n=
；n(n+1) 2
∴
=2n2+1 1+2+3+…+n
=4n2+2 n2+n
；4+ 2 n2 1+ 1 n
∴lim n→∞
=2n2+1 1+2+…+n lim n→∞
=4．4+ 2 n2 1+ 1 n
故答案为4．
搜题请搜索小程序“爱下问搜题”
计算
|
∵1+2+3+…+n=
；n(n+1) 2
∴
=2n2+1 1+2+3+…+n
=4n2+2 n2+n
；4+ 2 n2 1+ 1 n
∴lim n→∞
=2n2+1 1+2+…+n lim n→∞
=4．4+ 2 n2 1+ 1 n
故答案为4．