數學歸納法..........

對所有自數數n,n(n+1)(2n+1)能被3整除
當 n = 1
T(1) = n(n+1)(n+2) = 1x(1+1)(1+2) = 6
能被 3 整除
所以 n = 1 時成立
設 n= k 時成立
則
T(k) = k(k+1)(k+2) = 3M
(M 為某正整數)
當 n = k+1
T(k+1) = (k + 1)(k + 2)(k+ 3)
= (k + 1)(k + 2)k + (k + 1)(k + 2)3
= 3M + 3(k + 1)(k + 2)
= 3[M + (k + 1)(k + 2)]
所以 T(k + 1) 亦為 3 的倍數
依數學歸納法性質，可以證明對任何自然數，n(n+1)(2n+1)能被 3 整除。

Here is an alternative solution:
n(n + 1)(2n + 1) = n(n + 1)(n + 2 + n - 1) = n(n + 1)(n + 2) + (n - 1)n(n + 1)

Note that both n(n + 1)(n + 2) and (n - 1)n(n + 1) are the products of
3 consecutive integers, of which exactly one of them must be divisible by 3

so we may write n(n + 1)(n + 2) = 3M, (n - 1)n(n + 1) = 3K for some integers M and K

Hence we have n(n + 1)(2n + 1) = 3(M + K), which means n(n + 1)(2n + 1) is divisible by 3

Note: This solution does not require mathematical induction