化简:根号下n(n+1)(n+2)(n+3)+1
来源:学生作业帮助网 编辑:作业帮 时间:2024/11/24 17:56:05
x){3{f=[i';4
506I*'N~
fEQ`PSgm 2/{"~Ou tO?llm#K5ڍvbH @-v` $ف Q
化简:根号下n(n+1)(n+2)(n+3)+1
化简:根号下n(n+1)(n+2)(n+3)+1
化简:根号下n(n+1)(n+2)(n+3)+1
根号下n(n+1)(n+2)(n+3)+1
=根号下n(n+3)(n+1)(n+2)+1
=(n^2+3n)(n^2+3n+2)+1
设(n^2+3n)为k,
原式=根号下k(k+2)+1
=根号下(k^2+2k+1)
=根号下(k+1)^2
=k+1=n^2+3n+1
化简:根号下n(n+1)(n+2)(n+3)+1
根号下n(n+2)+1= n为自然数
求limn->无穷1/n(根号下1/n+根号下2/n+.+根号下n/n)
根号下n(n+1)(n+2)(n+3)+1 的化简怎样化简 根号下n(n+1)(n+2)(n+3)+1
根号(n+1)+n
根号【(2n+1)/(n²+n)】平方-4/(n平方+n )化简
求极限 n趋向于无穷 lim((根号下n^2+1)/(n+1))^n
求极限lim(n趋向于无穷)(n+1)(根号下(n^2+1)-n)
证明lim(n→∞){n-根号下n^2-n}=1/2
怎样化简根号下n(n+1)(n+2)(n+3)
lim(n→∞) 根号n+2-根号n+1/根号n+1-根号n
(-1)^n×1/根号下n(n+1)敛散性
已知n为正整数,求证:根号下n^2+n
根号下n^2+1-n=1/根号下n^2+1+n怎么算出来的
根号下n^2+n减去根号下n^2-2n的极限
lim n趋向于无穷大,n[(根号下n平方+1)-(根号下n平方-1)]
求极限 根号下(n方+n+1)-根号下(n方-n+1)
求极限n趋向于无穷 [(√n+2)-(√n+1)]√n Ps:是根号下的(n+2) 根号下的(n+1)