Prove by PMI,n(n+1)(2n+1)is divisible by 6

n(n+1)(2n+1) = 6, divisible by 6.

Let the result be true for n=k

Then, k(k+1)(2k+1) is divisible by 6.

So k(k+1)(2k+1) =6m (1)

Now to prove that the result is true for n=k+1

That is to prove, (K+1)(k+2)(2k+3) is divisible by 6.

(K+1)(k+2)(2k+3)=(k+1)k(2k+3)+(k+1)2(2k+3)=(k+1)k(2k+1)+(k+1)k2+(k+1)2(2k+3)

=6m+2(k+1)(k+2k+3) using (1)

=6m+2(k+1)(3k+3)

=6m +6(k+1)(k+1)=6[m+(k+1)^{2}]

So divisible by 6.

Hence, by PMI, the result is true for all n belong to R.