Key Concepts and Formulas

• Sum of an Infinite Geometric Series: The sum $S$ of an infinite geometric series with first term $a$ and common ratio $r$ is given by $S = \frac{a}{1-r}$, provided that $|r| < 1$.
• Algebraic Manipulation of Series: Infinite series can sometimes be rearranged and grouped to reveal simpler structures, such as geometric series.
• Co-prime Numbers: Two integers are co-prime if their greatest common divisor (GCD) is 1.

Step-by-Step Solution

1. Analyze the Structure of the Series: The given series is: $S = \left(\frac{1}{2}-\frac{1}{3}\right)+\left(\frac{1}{2^{2}}-\frac{1}{2 \cdot 3}+\frac{1}{3^{2}}\right)+\left(\frac{1}{2^{3}}-\frac{1}{2^{2} \cdot 3}+\frac{1}{2 \cdot 3^{2}}-\frac{1}{3^{3}}\right)+\left(\frac{1}{2^{4}}-\frac{1}{2^{3} \cdot 3}+\frac{1}{2^{2} \cdot 3^{2}}-\frac{1}{2 \cdot 3^{3}}+\frac{1}{3^{4}}\right)+\ldots$ Let's examine the $n$-th term (starting from $n=1$). The $n$-th term appears to be a sum of terms involving powers of $\frac{1}{2}$ and $\frac{1}{3}$. Specifically, for the $k$-th term within the $n$-th parenthesis (where $k$ ranges from 0 to $n$), the term is of the form $(-1)^k \frac{1}{2^{n-k} 3^k}$. For $n=1$: $\left(\frac{1}{2^1 3^0} - \frac{1}{2^0 3^1}\right)$ For $n=2$: $\left(\frac{1}{2^2 3^0} - \frac{1}{2^1 3^1} + \frac{1}{2^0 3^2}\right)$ For $n=3$: $\left(\frac{1}{2^3 3^0} - \frac{1}{2^2 3^1} + \frac{1}{2^1 3^2} - \frac{1}{2^0 3^3}\right)$ This suggests that the $n$-th term is the binomial expansion of $\left(\frac{1}{2} - \frac{1}{3}\right)^n$, but with alternating signs within the parenthesis. Let's rewrite the terms more carefully.

   The $n$-th group of terms (starting with $n=1$) is: $\sum_{k=0}^{n-1} (-1)^k \frac{1}{2^{n-k} 3^k}$ This is not quite correct. Let's look at the structure again: Term 1: $\frac{1}{2} - \frac{1}{3}$ Term 2: $\frac{1}{2^2} - \frac{1}{2 \cdot 3} + \frac{1}{3^2}$ Term 3: $\frac{1}{2^3} - \frac{1}{2^2 \cdot 3} + \frac{1}{2 \cdot 3^2} - \frac{1}{3^3}$ Term 4: $\frac{1}{2^4} - \frac{1}{2^3 \cdot 3} + \frac{1}{2^2 \cdot 3^2} - \frac{1}{2 \cdot 3^3} + \frac{1}{3^4}$

   The $n$-th parenthesis contains $n+1$ terms. The general term in the $n$-th parenthesis (starting $n=1$ for the first parenthesis) is of the form $\frac{1}{2^{n-k} 3^k}$ with an alternating sign. More precisely, for the $m$-th term in the series (where $m=1$ for the first term), let's consider the powers of 2 and 3.

   A more effective approach is to rearrange the entire series by grouping terms with common denominators involving powers of 2 and 3.

2. Rearrange and Group Terms: Let's expand the series and group terms based on the powers of $\frac{1}{2}$ and $\frac{1}{3}$. $S = \left(\frac{1}{2} - \frac{1}{3}\right) + \left(\frac{1}{2^2} - \frac{1}{2 \cdot 3} + \frac{1}{3^2}\right) + \left(\frac{1}{2^3} - \frac{1}{2^2 \cdot 3} + \frac{1}{2 \cdot 3^2} - \frac{1}{3^3}\right) + \left(\frac{1}{2^4} - \frac{1}{2^3 \cdot 3} + \frac{1}{2^2 \cdot 3^2} - \frac{1}{2 \cdot 3^3} + \frac{1}{3^4}\right) + \ldots$ We can rewrite this by collecting terms that are powers of $\frac{1}{2}$, terms that have one factor of 3, terms with two factors of 3, and so on.

   Terms with only powers of $\frac{1}{2}$: $\frac{1}{2} + \frac{1}{2^2} + \frac{1}{2^3} + \frac{1}{2^4} + \ldots$ Terms with one factor of 3 in the denominator: $-\frac{1}{3} - \frac{1}{2 \cdot 3} - \frac{1}{2^2 \cdot 3} - \frac{1}{2^3 \cdot 3} - \ldots$ Terms with two factors of 3 in the denominator: $\frac{1}{3^2} + \frac{1}{2 \cdot 3^2} + \frac{1}{2^2 \cdot 3^2} + \ldots$ Terms with three factors of 3 in the denominator: $-\frac{1}{3^3} - \frac{1}{2 \cdot 3^3} - \frac{1}{2^2 \cdot 3^3} - \ldots$

   So, the series can be rewritten as: $S = \left(\frac{1}{2} + \frac{1}{2^2} + \frac{1}{2^3} + \ldots\right) + \left(-\frac{1}{3} - \frac{1}{2 \cdot 3} - \frac{1}{2^2 \cdot 3} - \ldots\right) + \left(\frac{1}{3^2} + \frac{1}{2 \cdot 3^2} + \frac{1}{2^2 \cdot 3^2} + \ldots\right) + \left(-\frac{1}{3^3} - \frac{1}{2 \cdot 3^3} - \frac{1}{2^2 \cdot 3^3} - \ldots\right) + \ldots$

3. Factor and Identify Geometric Series: Now, let's factor out common terms from each group: $S = \left(\frac{1}{2} + \frac{1}{2^2} + \frac{1}{2^3} + \ldots\right) - \frac{1}{3}\left(1 + \frac{1}{2} + \frac{1}{2^2} + \ldots\right) + \frac{1}{3^2}\left(1 + \frac{1}{2} + \frac{1}{2^2} + \ldots\right) - \frac{1}{3^3}\left(1 + \frac{1}{2} + \frac{1}{2^2} + \ldots\right) + \ldots$ The expression in the parentheses, $1 + \frac{1}{2} + \frac{1}{2^2} + \ldots$, is an infinite geometric series with first term $a=1$ and common ratio $r=\frac{1}{2}$. Since $|r| < 1$, its sum is $\frac{1}{1 - \frac{1}{2}} = \frac{1}{\frac{1}{2}} = 2$.

   Substituting this sum back into the expression for $S$: $S = \left(\frac{1}{2} + \frac{1}{2^2} + \frac{1}{2^3} + \ldots\right) - \frac{1}{3}(2) + \frac{1}{3^2}(2) - \frac{1}{3^3}(2) + \ldots$ The first part, $\frac{1}{2} + \frac{1}{2^2} + \frac{1}{2^3} + \ldots$, is also an infinite geometric series with first term $a=\frac{1}{2}$ and common ratio $r=\frac{1}{2}$. Its sum is $\frac{\frac{1}{2}}{1 - \frac{1}{2}} = \frac{\frac{1}{2}}{\frac{1}{2}} = 1$.

   So, the expression for $S$ becomes: $S = 1 - \frac{2}{3} + \frac{2}{3^2} - \frac{2}{3^3} + \ldots$ $S = 1 + 2\left(-\frac{1}{3} + \frac{1}{3^2} - \frac{1}{3^3} + \ldots\right)$

4. Sum the Remaining Geometric Series: The series inside the parentheses, $-\frac{1}{3} + \frac{1}{3^2} - \frac{1}{3^3} + \ldots$, is an infinite geometric series with first term $a = -\frac{1}{3}$ and common ratio $r = -\frac{1}{3}$. Since $|r| < 1$, its sum is: $\frac{-\frac{1}{3}}{1 - \left(-\frac{1}{3}\right)} = \frac{-\frac{1}{3}}{1 + \frac{1}{3}} = \frac{-\frac{1}{3}}{\frac{4}{3}} = -\frac{1}{4}$

   Now, substitute this sum back into the expression for $S$: $S = 1 + 2\left(-\frac{1}{4}\right) = 1 - \frac{2}{4} = 1 - \frac{1}{2} = \frac{1}{2}$

5. Determine $\alpha$ and $\beta$, and Calculate $\alpha + 3\beta$: The sum of the series is $\frac{1}{2}$. We are given that the sum is $\frac{\alpha}{\beta}$, where $\alpha$ and $\beta$ are co-prime. So, $\alpha = 1$ and $\beta = 2$. The numbers 1 and 2 are co-prime since their greatest common divisor is 1. We need to calculate $\alpha + 3\beta$. $\alpha + 3\beta = 1 + 3(2) = 1 + 6 = 7$

Common Mistakes & Tips

• Incorrectly identifying the general term: The structure of the terms within each parenthesis can be confusing. Rearranging and grouping by powers of the bases is a more robust method.
• Forgetting the $|r|<1$ condition: Always verify that the common ratio of an infinite geometric series is less than 1 in absolute value before applying the sum formula. In this problem, all common ratios encountered are indeed less than 1.
• Sign errors: Carefully track the signs of terms, especially when identifying the common ratio in alternating series.

Summary

The given infinite series was first rearranged by grouping terms with common powers of 2 and 3 in the denominator. This revealed that the series could be expressed as a sum and difference of several infinite geometric series. By applying the formula for the sum of an infinite geometric series, $S = \frac{a}{1-r}$, to each component series, we found the total sum to be $\frac{1}{2}$. With $\alpha=1$ and $\beta=2$ (which are co-prime), the required value $\alpha + 3\beta$ was calculated.

The final answer is $\boxed{7}$.