求数列(2n^2+n+3)/(3n^2+4n+1)的极限如题

来源：学生作业帮助网编辑：六六作业网时间：2024/12/22 14:41:35

求数列(2n^2+n+3)/(3n^2+4n+1)的极限如题求数列(2n^2+n+3)/(3n^2+4n+1)的极限如题求数列(2n^2+n+3)/(3n^2+4n+1)的极限如题答案是2/3(2n^

求数列(2n^2+n+3)/(3n^2+4n+1)的极限如题
求数列(2n^2+n+3)/(3n^2+4n+1)的极限
如题

求数列(2n^2+n+3)/(3n^2+4n+1)的极限如题
答案是2/3

(2n^2+n+3)/(3n^2+4n+1)
=2/3

lim(2n^2+n+3)/(3n^2+4n+1)丨n=∞
=lim(2+1/n+3/n^2)/(3+4/n+1/n^2)丨n=∞
=(2+lim1/n+lim3/n^2)/(3+lim4/n+lim1/n^2丨n=∞
=(2+0+0)/(3+0+0)
=2/3。

数列a(n)=n (n+1)(n+2)(n+3),求S(n) 数列a(n)=n (n+1)(n+2)(n+3), 求S(n)怎么用高中数列原理解答? lim n →∞ (1^n+3^n+2^n)^1/n,求数列极限 1.求数列{n*2^n-n}的前n项和2.求数列{2n-3/2^（n-3）}的前n项和求数列 [（-2）^n+3^n]/[(-2)^(n+1)+3^(n+1)]的极限数列bn=2^n(4n-3),求Sn 求数列{n/2^n}的和S 求数列{（2n+3）-X^n}的和Tn 求n项和数列极限,通式为i/(i+n)就是n->无穷,1/(n+1)+2/(n+2)+3/(n+3)+...+n/2n; 数列n+(n^2+n^3)^(1/3)的极限已知b(n)=3/(2n+1)*(2n-1)求数列{b(n)}前n项的和已知an=5n(n+1)(n+2)(n+3),求数列{an}的前n项和Sn 求数列an=n(n+1) 的前n项和到 an=n(n+1)=[n(n+1)(n+2)-(n-1)n(n+1)]/3（裂项）数列bn=3^n/(3^n+2)[3^(n+1)+2],求Tn 求数列1/2,2/4,3/8...n/n^2的前n项和求数列{2的n次方分之1+3n-2}前n项的和. 求数列{(2n-1)*3^n}的前n项和求数列4,9,16,.,3n-1+2^n,.前n项的和Sn 求数列2n+1/3^n的前n项和