Q: 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!

A: That is a geometric sum:
$${ 2 }^{ 0 }+{ 2 }^{ 1 }+{ 2 }^{ 2 }+{ 2 }^{ 3 }+...+{ 2 }^{ n }=\frac{2^{n+1}-1}{2-1}$$
It is taught at highschools.

A: There is no need to estimate, the exact answer is $2^{n+1}-1$.
This is basic algebra as taught in schools.

A: 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 $a = 2$:
$$
\frac{2^{n+1} - 1}{2-1} = \sum\limits_{i = 1}^n {2^i} = 2^0+2^1+2^2+2^3+...+2^n
$$

A: Two great math courses that deal with sums and sequences such as the one you have defined are Real Analysis and basic Single-Variable Calculus. This problem specifically deals with geometric progression. Yes, you do learn some in high school, but not that much. Real Analysis is a subject that gives you a more structured intuition for these types of problems.
The solution to your problem is this by a geometric sum:
$$2^0+2^1+2^2+2^3+\cdot\cdot\cdot+2^n=\frac{2^{n+1}-1}{2-1}=\boxed{2^{n+1}-1}.$$

A: A more general answer for the sum of any infinite geometric series would be:
$S(x) = \frac{a(1  −r^X)}{(1  −r)}$
where x is the number of terms (x for the $2^n$ position is n+1), a is the first term ($2^0$) of the series, and r (r ≠ 1) is the common ratio of the terms (2/1 = 4/2 = ... = 2).
So for $2^0 + 2^1 + 2^2 + 2^3 +2^4$ it would be $S(5) = \frac{1*(1 - 2^5)}{(1 - 2)} = \frac{-31}{-1} = 31$

A: late to the party but i think it's useful to have a way of getting to the general formula.
this is a geometric serie which means it's the sum of a geometric sequence (a fancy word for a sequence where each successive term is the previous term times a fixed number). we can find a general formula for geometric series following the logic below
$$
a = \text{firstterm}\\
r = \text{common ratio}\\
n = \text{number of terms}\\
S_n = \text{sum of first n terms}\\
S_n = a + ar + ar^2 + \dots+ar^{n-1}\\
\\
\\
-rS_n = -ar-ar^2-ar^3 - \dots-ar^{n}\\
S_n-rS_n = a + ar + ar^2 + \dots+ar^{n-1}-ar-ar^2-ar^3 - \dots-ar^{n}\\
S_n-rS_n =a-ar^n = a(1-r^n)\\
S_n(1-r) = a(1-r^n)\\
S_n = \frac{a(1-r^n)}{(1-r)}
$$
then using a = 2
$$
S_n = \frac{1(1-2^n)}{(1-2)} = 2^n-1
$$
