數學知識交流---求值

1n + 2(n-1) + 3(n-2) + 4(n-3) + ... + (n-1)(2) + (n)(1) = (n+1 - 1) + 2(n+1 - 2) + 3(n+1 - 3) + ... + (n-1)(n+1 - (n-1)) + n(n+1 - n)= (n+1)(1 + 2 + 3 + ... + n) - (1 + 2² + 3² + ... + n²)= (n+1) n(n+1) / 2 - n(n+1)(2n+1) / 6= n (n+1) ( (n+1) / 2 - (2n+1)/6 )= n (n+1) ( 3n+3 - 2n -1 ) / 6= n (n+1) (n+2) / 6
1) 79 (79+1) (79+2) / 6 = 65320
2)(x/2) (x/2 + 1) (x/2 + 2) / 6= x (x + 2) (x + 4) / 12
3)1028 (1028 + 1) (1028 + 2) / 6 = 181591060

2011-07-29 22:51:28 補充：
Corrections :

1)

79 (79+1) (79+2) / 6 = 85320