# 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.

Neat!