00:01
We want to show that n -cube minus n is divisible by six for all n in the positive integers.
00:06
So we want to check the base case, which is n equals 1, since we're talking about the positive integers, then 1 -1 -1 -is equal to 0, which is divisible by 6.
00:21
So n -equals 1 is true.
00:25
We assume this is true up to n -equals k.
00:32
That would tell us k cubed minus k is divisible by six.
00:42
Then for n equals k plus 1, that would mean k plus 1 cubed minus k plus 1.
00:54
And we want to try to show that this thing is also divisible by 6.
00:59
We'll have to expand it.
01:01
So i'm going to take out a common factor of k plus 1.
01:04
And left with k plus 1 squared minus 1.
01:10
And that works out to k plus 1 times k squared plus 2k.
01:17
So this is easy to distribute.
01:19
This is k cubed plus 3k squared plus 2k.
01:30
And we want to find a way to use that inductive step.
01:33
So i want k cubed minus k somewhere.
01:35
So i will write it as k cubed minus k.
01:37
Plus 3k squared, which would change my last term to 3k.
01:43
So 3k, negative k, that creates the negative 2k in the previous step.
01:48
So this is still equivalent.
01:51
The k -kub minus k here can be written as 6a because it is divisible by 6...