Consider the set of all fractions of the form $\frac{2^m}{3^n},$ where $m$ and $n$ are integers such that $0 \le m \le n.$ For example, $\frac{1}{9}$ and $\frac{8}{27}$ would be included in this set, but not $\frac{32}{81}.$ What is the sum of all the numbers in this set?

(A) $\frac{81}{64}$ (B) $\frac{27}{16}$ (C) $\frac{9}{4}$ (D) $\frac{27}{8}$ (E) $\frac{9}{2}$

We are tasked with finding the sum of all fractions of the form $ \frac{2^m}{3^n} $, where $ m $ and $ n $ are integers such that $ 0 \leq m \leq n $.

### Step 1: List the Fractions
The fractions take the form:

\[
\frac{2^m}{3^n} = 2^m \cdot 3^{-n}
\]

where $ m $ and $ n $ are constrained by $ 0 \leq m \leq n $. For each value of $ n $, $ m $ ranges from 0 to $ n $, so we can start summing the possible fractions for each $ n $.

### Step 2: Set up the Sum for Each $ n $
For each $ n $, the possible values of $ m $ range from 0 to $ n $, and the sum of the fractions for a fixed $ n $ is:

\[
\sum_{m=0}^n \frac{2^m}{3^n} = \frac{1}{3^n} + \frac{2}{3^n} + \frac{2^2}{3^n} + \cdots + \frac{2^n}{3^n}
\]

This is a geometric series with first term $ 1 $ and common ratio $ 2 $, summing over $ n+1 $ terms:

\[
\sum_{m=0}^n \frac{2^m}{3^n} = \frac{1}{3^n} \left( 1 + 2 + 2^2 + \cdots + 2^n \right)
\]

The sum of the geometric series $ 1 + 2 + 2^2 + \cdots + 2^n $ is:

\[
\frac{2^{n+1} - 1}{2 - 1} = 2^{n+1} - 1
\]

Thus, the sum for a fixed $ n $ becomes:

\[
\sum_{m=0}^n \frac{2^m}{3^n} = \frac{1}{3^n} \cdot (2^{n+1} - 1) = \frac{2^{n+1} - 1}{3^n}
\]

### Step 3: Set up the Total Sum
The total sum is the sum of these series over all values of $ n \geq 0 $:

\[
\text{Total sum} = \sum_{n=0}^{\infty} \frac{2^{n+1} - 1}{3^n}
\]

This can be broken into two simpler sums:

\[
\text{Total sum} = \sum_{n=0}^{\infty} \frac{2^{n+1}}{3^n} - \sum_{n=0}^{\infty} \frac{1}{3^n}
\]

### Step 4: Solve Each Sum

#### First Sum: $ \sum_{n=0}^{\infty} \frac{2^{n+1}}{3^n} $
This is a geometric series with first term $ 2 $ and common ratio $ \frac{2}{3} $:

\[
\sum_{n=0}^{\infty} \frac{2^{n+1}}{3^n} = 2 \sum_{n=0}^{\infty} \left( \frac{2}{3} \right)^n = 2 \cdot \frac{1}{1 - \frac{2}{3}} = 2 \cdot 3 = 6
\]

#### Second Sum: $ \sum_{n=0}^{\infty} \frac{1}{3^n} $
This is a geometric series with first term $ 1 $ and common ratio $ \frac{1}{3} $:

\[
\sum_{n=0}^{\infty} \frac{1}{3^n} = \frac{1}{1 - \frac{1}{3}} = \frac{3}{2}
\]

### Step 5: Final Calculation
The total sum is:

\[
\text{Total sum} = 6 - \frac{3}{2} = \frac{12}{2} - \frac{3}{2} = \frac{9}{2}
\]

Thus, the sum of all the numbers in this set is $ \boxed{\frac{9}{2}} $.

Would you like any additional details or clarifications? 

Here are five related questions for further exploration:
1. How do geometric series properties simplify infinite summations?
2. Can this approach be applied to similar sums with different powers of primes?
3. What if the set included fractions with different base exponents?
4. How does this sum relate to the behavior of infinite series?
5. How does the constraint $ 0 \leq m \leq n $ affect the sum?

Tip: When summing series with multiple terms, breaking them into simpler, geometric series can make computation much easier!

Consider the set of all fractions of the form $ \frac{2^m}{3^n} $, where $ m $ and $ n $ are integers such that $ 0 \le m \le n. $ For example, $ \frac{1}{9} $ and $ \frac{8}{27} $ would be included in this set, but not $ \frac{32}{81}. $ What is the sum of all the numbers in this set?

(A) $ \frac{81}{64} $ (B) $ \frac{27}{16} $ (C) $ \frac{9}{4} $ (D) $ \frac{27}{8} $ (E) $ \frac{9}{2} $

Explore how to find the sum of all fractions of the form $ \frac{2^m}{3^n} $, where $ m \leq n. $ This problem involves geometric series and summing infinite fractions. Includes detailed steps and final answer: $ \boxed{\frac{9}{2}} $. Ideal for advanced high school and undergraduate students.

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