00:01
Okay, so let's do this.
00:05
So we have d.
00:14
So we have d n is equal to 3 to the n minus 2 to the n, right? and then we have we want to show it also is equal to that it's satisfied the recurrence.
00:30
I guess i'll do it in terms of n.
00:32
Dn is equal to 5 times d to d to the n minus 1 plus.
00:38
Plus six or not plus minus minus six times d to the n minus two.
00:49
Okay, so let's just plug in.
00:52
So what is d to the n minus one? so this is five times three to the n minus one, minus two to the n minus one.
01:10
And now i'm also going to split this up as five of these, 3 to the n minus 2 minus 2 to the n minus 2 and then also one more so 3 to the n minus 2 so you get a minus so this is plus 2 to the end minus 2 okay so let's do this so this is going to be 5 times so we get 3 to the n minus 1 minus 3 to the n minus 2 and then we get plus 2 to the n minus 2 minus 2 to the n minus 1.
02:04
Okay, now these ones we can compute.
02:08
So if we, for example, for this one, we can bring out a 3 to the n minus 2, and we get times 3 minus 1...