吉米多维亚:求证Σ(from 1 to n)k=n(n+1)/2
Σ(from 1 to n)k²=n(n+1)(2n+1)/6
卓里奇:已知Σ(from 1 to n)k=n(n+1)/2
Σ(from 1 to n)k²=n(n+1)(2n+1)/6
求Σ(from 1 to n)k^m的通解
Σ(from 1 to n)k²=n(n+1)(2n+1)/6
卓里奇:已知Σ(from 1 to n)k=n(n+1)/2
Σ(from 1 to n)k²=n(n+1)(2n+1)/6
求Σ(from 1 to n)k^m的通解