Sum of a series of a number raised to incrementing powers

Question

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!

Answer 1

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:

$$2^0+2^1+2^2+2^3+\cdot\cdot\cdot+2^n=\frac{2^{n+1}-1}{2-1}=\boxed{2^{n+1}-1}.$$

Answer 2

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.

Answer 3

There is no need to estimate, the exact answer is $2^{n+1}-1$.

This is basic algebra as taught in schools.

Answer 4

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

Answer 5

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 $S(5) = \frac{1*(1 - 2^5)}{(1 - 2)} = \frac{-31}{-1} = 31$

Answer 6

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

$$ S_n = \frac{1(1-2^n)}{(1-2)} = 2^n-1 $$