The next step is to assume the proposition is true for n.
The task is to show that if the proposition is true for n, then it would be true for n+1. Once that is established, then the proof is complete.
6|n^3+5n => 6|1+5 is true for n=1
Assume 6|N^3+5 is true for n, then
for n+1
(n+1)^3+5(n+1)
=n^3+5n +3n^2+3n+6
=n^3+5n +3n(n+1) + 6
We now examine the three terms:
n^3+5n is divisible by 6 by initial assumption.
6 is divisible by 6.
3n(n+1) falls into two cases:
1. n is odd, then n+1 is even, therefore 6 divides 3*(n+1)
2. n is even, then 6 divides 3n.
Since all three terms are divisible by 6, we only have to extract the factor of 6 from each term and declare the expression (n+1)^3+5(n+1) is also divisible by 6.
By the principle of mathematical induction, the proposition is proved. QED.