To prove a statement by induction, it has to be shown that the following two conditions are satisfied:
1) The statement is true for n = 1 (or any applicable initial value of n)
2) Assuming that the statement is true for n = k, the statement is ALSO true for n = k+ 1.
In this case, we need to prove that n(n+1)(n+2)(n+ 3) is divisible by 24. It is easy to check that the first condition is satisfied for n = 1:
n(n+ 1)(n+2)(n+3) = 1(2)(3)(4) = 24, which is divisible by itself.
Now let's see if can show that the second condition is also satisfied. Assume that for some n = k, k(k+1)(k+2)(k+3) is divisible by 24. We then need to show that if this is the case, then this expression for n = k+ 1 will also be divisible by 24. In other words, we need to show that
(k+1)(k+2)(k+3)(k+4) is also divisible by 24.
To do this, multiply out both expressions:
`k(k+1)(k+2)(k+3) = (k^2+ k)(k^2 + 5k + 6) = k^4 + 6k^3 + 11k^2 + 6k`
The second expression will be
`(k+1)(k+2)(k+3)(k+4) = (k^2 + 3k + 2)(k^2 + 7k + 12)`
`=k^4 + 10k^3 + 35k^2 + 50k + 24`
The next part is tricky. We are going to break up some of the terms of the second expression in the following way:
`10k^3 = 6k^3 + 4k^3`
`35k^2 = 11k^2 + 24k^2`
`50k = 6k + 44k`
Then the second expression can be rewritten as follows, by re-arranging the broken up terms:
`(k^4 + 6k^3 + 11k^2 + 6k) + (4k^3 + 24k^2 + 44k + 24)`
Notice that the expression in the first parenthesis is identical to the first expression, multiplied out k(k+1)(k+2)(k+3). This is divisible by 3 by assumption.
The second parenthesis contain terms 24k^2 and 24, which are clearly divisible by 24. We still need to show that the sum of remaining terms,
`4k^3 + 44k = 4(k^3 + 11k)` is divisible by 24. It would require the separate induction process to show that `k^3 + 11k` is in fact divisible by 6 (try it :), and then the product
`4(k^3 + 11k) ` is divisible by 24.
So we have shown that all parts of the sum constituting the second expression,
(k+1)(k+2)(k+3)(k+4) are divisible by 24, as long as k(k+1)(k+2)(k+3) is divisible by 24.
So, by induction, n(n+1)(n+2)(n+3) is divisible by 24.