1234door 寫到:
其實不難
Hint:
S1 = 1 + 2 + 3 + .. + n = n(n+1)/2
S2 = 12+ 22 + .. + n2 = n(n+1)(2n+1)/6 = n(n+1)/2 * [(2n+1)/3] = S1(2n+1)/3
S3 = 13+ 23 + .. + n3 = [(n+1)4-(n+1)]/4 -3S2/2 -S1 = n(n+1)(n2+3n+3)/4-3S2/2 -S1 = S1(n2+3n+3)/2-3S2/2 -S1 = S1的倍數
...
Sm = [(n+1)m+1-(n+1)] / (m+1) + S1的倍數
證明 [(n+1)m+1-(n+1)] / (m+1)是S1的倍數, 就留給其他人完成吧.