00:02
Okay, so we treat this one like any other problem before, just with fractions.
00:08
So our first term in the end equals 1 for our base case is 1 3rd, and that's going to be equal to 1 times 5 -6th times 1 minus 1 -half.
00:29
And now, if we combine our fractions, 1 -half goes to 3 -6th, and we're left with 2 -6th.
00:46
And then 2 .6th reduced is just one -third.
00:52
So our base case does work for this question.
01:00
Next we prove our inductive case.
01:04
So if we let n equal to k, we get 1 third plus 2, plus all the way to 5 thirds k minus 4 thirds.
01:30
And that's going to be equal for equation k times 5, 6k minus 1 .5.
01:51
Now we add in the next term, which is just however simple, we can leave it.
02:14
So our next term is going to be 5 thirds times k plus 1 minus 4 thirds.
02:31
That's all it needs to be because we know that's going to be our next term.
02:34
And then that's going to be equal to k times 5, 6k minus 1 half plus that same thing.
03:06
So we have our k plus 1 case on the top equation or the left -hand side.
03:12
So now we just have to prove it for our right equation.
03:16
So now what i'm going to do is i'm going to distribute everything...