Sum of a series of a number raised to incrementing powers How would I estimate the sum of a series of numbers like this:  $2^0+2^1+2^2+2^3+\cdots+2^n$.  What math course deals with this sort of calculation?  Thanks much!

Interestingly, this sort of summation is not too difficult to prove either.
If we swap $2$ for a generic number $a$:
$$
\text{(1) } \sum\limits_{i = 1}^n {a^i} = a^0+a^1+a^2+a^3+...+a^n
$$
Multiplying $(1)$ by $a$ gives:
$$
\text{(2) } a\sum\limits_{i = 1}^n {a^i} = a^1+a^2+a^3+a^4+...+a^{n+1}
$$
Subtracting $(1)$ from $(2)$:
$$
\text{(3) } a\sum\limits_{i = 1}^n {a^i} - \sum\limits_{i = 1}^n {a^i} == (a-1)\sum\limits_{i = 1}^n {a^i}
$$
From $(1)$ and $(2)$ we can see this is the same as:
$$
\text{(4) } a^1+a^2+a^3+a^4+...+a^{n+1} - (a^0+a^1+a^2+a^3+...+a^n) = (a-1)\sum\limits_{i = 1}^n {a^i}
$$
Since all the middle bits cancel each other out, this simply leaves:
$$
\text{(5) } a^{n+1} - a^0 = (a-1)\sum\limits_{i = 1}^n {a^i}
$$
And since $a^0$ is always equal to $1$:
$$
\text{(6) } a^{n+1} - 1 = (a-1)\sum\limits_{i = 1}^n {a^i}
$$
Rearranging:
$$
\text{(7) } \frac{a^{n+1} - 1}{a-1} = \sum\limits_{i = 1}^n {a^i}
$$
And returning to the original post, where $a = 2$:
$$
\frac{2^{n+1} - 1}{2-1} = \sum\limits_{i = 1}^n {2^i} = 2^0+2^1+2^2+2^3+...+2^n
$$
