Sum of the Powers of $2$ Suppose I have a sequence consisting of the first, say, $8$ consecutive powers of $2$ also including $1$: $1,2,4,8,16,32,64,128$. Why is it that for example, $1 + 2 + 4 = 7$ is $1$ less than the next term in the series, $8$? Even if one was to try, for instance, $524,288 + 1,048,576$ ($2^{19}$ and $2^{20}$), the next term ($2^{21}$) will be $2,097,152$, which is one greater than the previous sum. I found this whilst solving the "if you are given a penny each day and it doubles" problem. Can someone please explain this to me? Thank you.

If you think about the list of the positive integers written in base 2, it goes like this:
$$\begin{align*}
1&=1\cdot 2^0 &=1_2\\
2&=1\cdot 2^1 + 0\cdot 2^0 &=\textbf{10}_2\\
3&=1\cdot 2^1+1\cdot 2^0 &=11_2\\
4&=1\cdot 2^2+0\cdot 2^1+0\cdot 2^0 &=\textbf{100}_2\\
5&=1\cdot 2^2+0\cdot 2^1+1\cdot 2^0 &=101_2\\
6&=1\cdot 2^2+1\cdot 2^1+0\cdot 2^0 &=110_2\\
7&=1\cdot 2^2+1\cdot 2^1+1\cdot 2^0 &=111_2\\
8&=1\cdot 2^3+0\cdot 2^2+0\cdot 2^1+0\cdot 2^0 &=\textbf{1000}_2\\
9&=1\cdot 2^3+0\cdot 2^2+0\cdot 2^1+1\cdot 2^0 &=1001_2\\
\vdots&=\vdots &\vdots\\
2^{n}-1&=1\cdot2^{n-1}+1\cdot 2^{n-2}+\cdots + 1\cdot 2^0 &=111\ldots 1_2\\
2^n&=1\cdot 2^n + 0\cdot 2^{n-1} + \cdots + 0\cdot 2^0 &=\textbf{100}\ldots\textbf{00}_2
\end{align*}$$
The number of digits in the expression in base 2 changes exactly at the powers of 2, and right before changing the number (in base 2) looks like a sequence of 1's, which corresponds to the sum $2^0+\ldots+2^{n-1}$ for some $n$. 
Note that a similar idea will work for any base $b$.
