LIM1+3+5+…+(2n+1)/n^+1

来源：百度知道编辑：UC知道时间：2024/06/17 23:13:43

1+3+5+…+(2n+1)
有n+1项
所以=(1+2n+1)(n+1)/2=(n+1)^2
所以LIM1+3+5+…+(2n+1)/(n^2+1)
=(n+1)^2/(n^2+1)
=(n^2+2n+1)/(n^2+1)上下同除n^2
=[1+2*(1/n)+1/n^2]/(1+1/n^2)
n→∞
则1/n→0,1/n^2→0
所以极限=(1+2*0+0)/(1+0)=1

lim(n->无穷){1+3+5+…+(2n+1)}/(n^+1)
=lim(n->无穷)n(2n+2)/(n^2+1)
=lim(n->无穷)4/2
=2

lim1^2+2^2+.....+n^2 已知m,n为正整数，求出满足等式3n+4n+5n+…+（n+2）n=(n+3)n的所有正整数n n×(n-1)×(n-1)求和,n为2、3、4……n 1×2×3+2×3×4+3×4×5+……n(n+1)(n+2) 1+3+5+… +(2n-5)+(2n-3)+(2n-1)+(2n+1) n+2*(n-1)+3*(n-2)+4*(n-3)+…… 求lim5^n-4^n-1/[(5^n+1)+3^n+2] 1×n+2(n-1)+3(n-2)+…(n-3)×4+(n-2)×3+(n-1)×2+n×1=? n.n+n-1=0则n.n.n-n.n+3n+5=？ 1*2+3*4+4*5+...+n(n-1)=?