@amusingminnow Great question! This is a pattern that works for all powers of two, actually.

Let's add them one at a time and see if there's anything you notice:

\(2^0=\boxed{1}\)

\(2^0+2^1=\boxed{3}\)

\(2^0+2^1+2^2=\boxed{7}\)

\(2^0+2^1+2^2+2^3=\boxed{15}\)

\(2^0+2^1+2^2+2^3+2^4=\boxed{31}\)

What's special about all of the answers is that they're all \(1\) less than the next power of \(2\)!

Knowing this is super helpful, it saves us from having to do a ton of addition. For example, we can use this shortcut when doing \(2^0+2^1+2^2+...+2^9\), because we know that this is equal to the next power of \(2,\) which is \(2^{10}=1024\), minus \(1\)!

So \(2^0+2^1+2^2+...+2^9=1023\).

If you really want to get into the nitty gritty of why this works, you have to factor. Let's use \(2^4-1\).

Remember difference of squares? We can use that here!

\(2^4-1=(2^2-1)(2^2+1)\)

And since \(2^2-1\) is a difference of two squares also, we can keep going.

\(2^4-1=(2^2-1)(2^2+1)=(2-1)(2+1)(2^2+1)\).

Well, \((2-1)\) is just \(1\), so the expression is equal to \((2+1)(2^2+1)\). Expand \((2+1)(2^2+1)\), and you'll get \(2^3+2^2+2+1=2^3+2^2+2^1+2^0\).

So,

\(2^3+2^2+2^1+2^0=2^4-1\).

You can use factoring to prove this for all of the powers of \(2\)