00:02
Okay, so we're trying to prove this geometric series.
00:08
So we have our base case, n equals 1.
00:16
The left -hand side for our first term is just 1, and that's going to be equal to 1 minus r to the first power, divided by 1 minus r, which is just 1 minus r over 1 minus r, which is 1.
00:33
So our base case works very well.
00:39
Now for our inductive case.
00:45
So given n equals k, 1 plus r, all r up to r to the k power, is equal to 1 minus r, excuse me, k minus 1 power.
01:08
Let me rewrite that.
01:12
R to the k minus 1 power is equal to 1 minus r to the k, divided by 1 minus 1...