Powers of Two

If you add the successive powers of two (starting at 2^{0}), you get one less than the next power of two. This is usually written as 1 + 2 + 2^{2} + ... + 2^{n-1} = 2^{n} - 1.

The first time I saw this, I thought to myself “Whoa, why does this happen?!” Better yet, I wanted to know “What does this idea look like?”


The picture below is usually used in a visual proof to show that \frac{1}{2} + \frac{1}{2^{2}} + \frac{1}{2^{3}} + \frac{1}{2^{4}} + ... converges to 1. The idea is that every cut produces two smaller, identical rectangles, and one of these rectangles can be cut into two even smaller, identical rectangles. Beginning with \frac{1}{2}, you add on \frac{1}{2} of a \frac{1}{2}, and then \frac{1}{2} of that \frac{1}{2} of a \frac{1}{2}, and so on. This can go on forever without ever exceeding the area of the square, which is equal to 1. Neat!


Let’s look at this picture differently and ask a different question about it.


Question 1: How many of the blue rectangle fits inside the bolded area?

Answer 1: If the blue rectangle is \frac{1}{2^{n}} of the entire square, then by definition there are 2^{n} blue rectangles in the entire square and 2^{n} - 1 blue rectangles in the bolded area. Done!

Question 2: How many of the blue rectangle fits inside the bolded area?

Answer 2: One blue rectangle fits inside one blue rectangle. Two blue rectangles fit inside the next largest. Four blue rectangles fit inside the next-next largest rectangle…

This keeps going until you ask, “How many blue rectangles fit inside the \frac{1}{2} piece?” Since 2^{n} blue rectangles fit inside the whole square, we can say that half as many fit inside the \frac{1}{2} piece. So 2^{n - 1} blue rectangles.


Let’s add up all the blue rectangles we’ve counted in each of the highlighted rectangles above:

1 + 2 + 4 + ... + 2^{n - 1}

The same counting question (how many blue squares fit inside the bold area?) can be answered in two different ways. So although each answer yields a different expression, both expressions must be equal.