M. I. Question

For n = 1 ,
1(1+1)(2*1 + 1) = 6 is divisible by 6.
Assume that when n = k ,
k(k+1)(2k+1) is divisible by 6 ,
Let k(k+1)(2k+1) = 6M ,
When n = k+1 ,
(k+1)(k+1 + 1)(2k+2 + 1)
= (k+1)(k+2)(2k+3)
= (k+1) (2k^2 + 4k + 3k + 6)
= (k+1) [(2k^2 + k) + (6k + 6)]
= (k+1) [ k(2k+1) + 6(k+1) ]
= k(k+1)(2k+1) + 6(k+1)^2
= 6M + 6(k+1)^2 which is divisible by 6.

當n=1時
1*(1+1)*(2*1+1)=6
So n=1時為真
設 n=k時為真
即存在正整數p使得
k(k+1)(2k+1)=6p
2k^3+3k^2+k=6p
2k^3=6p－3k^2－k
So
(k+1)*(k+1+1)*[2*(k+1)+1]
=(k+1)(k+2)(2k+3)
=(k^2+3k+2)(2k+3)
=2k^3+9k^2+13k+6
=6p-3k^2－k+9k^2+13k+6
=6p+6k^2+12k+6
=6(p+k^2+2k+1)
So n=k+1 時為真