Induction involves three steps. We'll follow each of them now...
1. PROVE IT'S TRUE FOR THE BASE CASE (n = 1)
n = 1 is the lowest possible value for n. So, plug that in and see if it makes the statement true:
n(n2 - 1)(n + 2)
(1)((1)2 - 1)((1) + 2)
(1)(1 - 1)(3)
0 is indeed divisible by 4. So, we've proven it's true in the base case.
2. ASSUME IT'S TRUE FOR THE CASE n - 1
That means we're going to assume that (n - 1)((n - 1)2 - 1)((n - 1) + 2) is divisible by 4. Let's make that expression a bit simpler...
(n - 1)(n2 - 2n + 1 - 1)(n + 1)
(n - 1)(n2 - 2n)(n + 1)
(n - 1)(n + 1)(n2 - 2n)
(n2 - n + n - 1)(n2 - 2n)
(n2 - 1)(n2 - 2n)
n4 - 2n3 - n2 + 2n
So, that expression is divisible by 4, by assumption.
3. PROVE IT'S TRUE FOR n, GIVEN THAT IT'S TRUE FOR n - 1
So, now we'd like to prove that n(n2 - 1)(n + 2) is divisible by 4. Let's simplify that a bit...
(n3 - n)(n + 2)
n4 + 2n3 - n2 - 2n
That looks a lot like something we've assumed is divisible by 4: n4 - 2n3 - n2 + 2n
We're missing the -2n3 and the +2n, though, so let's subtract 2n3 and add 2n. To make sure I haven't changed the expression's value at all, I'll also add 2n3 and subtract 2n elsewhere:
n4 - 2n3 + 2n3 + 2n3 - n2 + 2n - 2n - 2n
Those red terms cancel each other out, and the rest is just the same expression we had before. So, let's rearrange that a bit:
(n4 - 2n3 - n2 + 2n) + 2n3 + 2n3 - 2n - 2n
(n4 - 2n3 - n2 + 2n) + 4n3 - 4n
(n4 - 2n3 - n2 + 2n) + 4(n3 - n)
Our assumption says that the first parenthetical portion is definitely divisible by 4, since that's the n - 1 case. And the second portion is definitely divisible by 4, since it's a multiple of 4 (I could factor 4 out). So, the entire expression is indeed some multiple of 4 plus some multiple of 4, which is divisible by 4.
We've therefore proven it's true in the n case.
And that's the end of our proof! By induction, the expression is divisible by 4 for any n ≥ 1.
This works because we've done the following:
(a) We proved that, assuming it's true for one case, it's true for the next case (STEP 3)
(b) We proved that it's true for the first case (STEP 1)
So, since it's true for the first case and for any case that follows a true case, it's true for every case.