Question

大一高数求极限求极限 1/n^3 x [1^2+3^2+...+(2n-1)^2](1+5/n)^n给个过程 答案知道的

Answer 1

1^2+2^2+...n^2=n(n+1)(2n+3)/6

则：1^2+2^2+...(2n)^2=2n(2n+1)(4n+3)/6

[1^2+2^2+...(2n)^2]-[1^2+3^2+...+(2n-1)^2]

=2^2+4^2+6^2+...+(2n)^2=4[1^2+2^2+...n^2]=4n(n+1)(2n+3)/6

得：1^2+3^2+...+(2n-1)^2

=2n(2n+1)(4n+3)/6-4n(n+1)(2n+3)/6

=[n(8n^2+10n+3-4n^2-10n-6)]/3

=n(4n^2-3)/3

则：

1/n^3 x [1^2+3^2+...+(2n-1)^2]= n(4n^2-3)/(3 * n^3)

=(1/3)(4n^3-3n)/n^3=(1/3)*(4-3/n^2)

n->无穷 limSn=4/3

(1+5/n)^n=[(1+5/n)^(n/5)]^5

n->无穷 lim (1+5/n)^n=e^5