00:01
This problem, we want to show that n -cube minus n is divisible by 3 for all n in the positive integers.
00:06
So we check our base case.
00:09
Base case is n equals 1 because it's positive integers.
00:13
Well, that would mean 1 cubed minus 1 in the expression.
00:20
That'll equal to 0 is divisible by 3.
00:24
0 is divisible by anything.
00:27
So this is true, which means we can conclude n equals 1.
00:31
Is next assume true up to n equals k then we can say that k cubed minus k is divisible by three then when n is equal to k plus one we would have k plus one cubed minus k plus one and we would try to change this so that we can clearly tell that it is divisible by three.
01:16
So this can be written as a common factor of k plus one.
01:25
We can take it out.
01:26
So it'll be k plus 1 times k plus 1 squared minus 1.
01:35
We'll have k plus 1.
01:37
And this is k squared plus 2k plus 1.
01:40
So we just have k squared plus 2k left because of the minus 1 at the end...