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

Answer 2

[1^2+3^2+...+(2n-1)^2]
1^2+2^2+...n^2=n(n+1)(2n+1)/6
[1^2+3^2+...+(2n-1)^2]
=(2n-1)(2n-1+1)[2(2n-1)+1]/6
=(2n-1)2n*(4n-1)/6
1/n^3* [1^2+3^2+...+(2n-1)^2]=2*2*4/6=16/6=8/3

2.limn->无穷 (1+5/n)^n
=lim n->无穷 [(1+5/n)^(n/5)]^5
=e^5
希望对你有帮助

Answer 3

1^2+2^2+...n^2=n(n+1)(2n+1)/6
主要是运用这个公式