Let f(n) = n(n-1)(2n+5)
f(1) = 1(1-1)(2+5) = 0 Since zero is divisible by 6 the statement is proven to be divisible for n = 1
Assuming f(k) is divisible by 6 for all k which are positive integers greater than 1.
Therefore, f(k+1) = (k+1)(k)[2(k+1)+5]
f(k+1) - f(k) = (k+1)(k)[2(k+1)+5] - [k(k-1)(2k+5)]
= 2k3 +9k2 +7k - 2k3 -3k2 +5k
=6k2 + 12
= 6(k2 +2)
Hence, f(k+1) = 6(k2 +2) + f(k) is proven to be divisible when n=k+1
If f(n) is divisible when n=k it is shown to be divisible when n=k+1 where k is any positive integer greater than 1.