00:01
So with this problem, we're going to do a geometric sum.
00:04
And for the infinite geometric sum, what the formula is, is, well, sum to infinity, would be the first term divided by 1 minus the common ratio.
00:18
And so this is only applicable if the common ratio is less than 1.
00:24
So as we look at the problem, from k equals 1 to infinity of 8 times 1 4th to the k power, you can start to write out some of the problems, like 1 4th to the first power is 1 4th times 8 would give me 2.
00:43
And then if i switch to squared, i would have 1 16th times 8, which would be 1 half.
00:51
And maybe i'll just do one more, because there's a common ratio here would be 1 8th all the way to infinity.
00:58
And so what's the common ratio from one thing to the next is multiplied by 1 4th.
01:04
It's the base in that piece.
01:06
And that satisfies this condition, that r is smaller than 1.
01:11
So yes, the answer is a...