00:01
All right, so for this one, we need to apply a previous exercise.
00:05
And so the result from that one is that a .n is equal to s times fn minus 1 plus t times f of n.
00:28
Okay, and so what we're looking for here is we want to express it in the form a .n is equal to a m minus 1.
00:40
Plus a n minus 2 plus 1, right, where n is greater than or equal to 2, a not equals 0, and a 1 is equal to 1.
00:57
So we'll just rename, we'll set bn equal to a .n plus 1.
01:14
Okay.
01:16
So if bn is equal to a .n plus 1, then it's going to be equal to a n minus 1 plus a .n.
01:24
2 plus 1 plus 1, right? so we have plus 1 from a .n plus the 1 here.
01:29
So that's plus 2, which is going to be equal to a .n minus 1 plus 1 plus a .n minus 2 plus 1.
01:46
Well, these are really just bns, right? so this is equal to bn minus 1 plus bn minus 2...