If the sum of the series $$\left(\frac{1}{2} - \frac{1}{3}\right) + \left(\frac{1}{2^2} - \frac{1}{2 \cdot 3} + \frac{1}{3^2}\right) + \left(\frac{1}{2^3} - \frac{1}{2^2 \cdot 3} + \frac{1}{2 \cdot 3^2} - \frac{1}{3^3}\right) + \left(\frac{1}{2^4} - \frac{1}{2^3 \cdot 3} + \frac{1}{2^2 \cdot 3^2} - \frac{1}{2 \cdot 3^3} + \frac{1}{3^4}\right) + \ldots$$ is $$\frac{\alpha}{\beta}$$, where $$\alpha$$ and $$\beta$$ are co-prime, then $$\alpha + 3\beta$$ is equal to _____.

$$x^{k+1} - y^{k+1} = (x - y)(x^k + x^{k-1}y + \dots + y^k)$$

Let $$a = \frac{1}{2}$$ and $$b = \frac{1}{3}$$. The given series $$S$$ can be written as:

$$S = (a - b) + (a^2 - ab + b^2) + (a^3 - a^2b + ab^2 - b^3) + \dots$$

$$T_k = a^k - a^{k-1}b + a^{k-2}b^2 - \dots + (-1)^{k-1}b^k$$

$$(a + b)T_k = (a + b)(a^k - a^{k-1}b + \dots + (-1)^{k-1}b^k)$$

This matches the expansion of $$a^{k+1} - (-b)^{k+1}$$:

$$(a + b)T_k = a^{k+1} - (-b)^{k+1}$$

$$(a + b)S = \sum_{k=1}^{\infty} a^{k+1} - \sum_{k=1}^{\infty} (-b)^{k+1}$$

This results in two infinite geometric progressions.
The first GP has a first term of $$a^2$$ and a common ratio of $$a$$.
The second GP has a first term of $$(-b)^2 = b^2$$ and a common ratio of $$-b$$.

$$(a + b)S = \frac{a^2}{1 - a} - \frac{b^2}{1 - (-b)}$$

$$\left(\frac{1}{2} + \frac{1}{3}\right)S = \frac{\left(\frac{1}{2}\right)^2}{1 - \frac{1}{2}} - \frac{\left(\frac{1}{3}\right)^2}{1 + \frac{1}{3}}$$

$$S = \frac{5}{12} \times \frac{6}{5} = \frac{1}{2}$$

$$\alpha + 3\beta = 1 + 3(2) = 1 + 6 = 7$$