Let $$S = \{\alpha : \log_2(9^{2\alpha-4} + 13) - \log_2(\frac{5}{2} \cdot 3^{2\alpha-4} + 1) = 2\}$$. Then the maximum value of $$\beta$$ for which the equation $$x^2 - 2(\sum_{\alpha \in s} \alpha)^2 x + \sum_{\alpha \in s}(\alpha+1)^2\beta = 0$$ has real roots, is _____.

We need to find the set $$S$$ and then determine the maximum value of $$\beta$$.

$$\log_2\left(9^{2\alpha-4} + 13\right) - \log_2\left(\frac{5}{2} \cdot 3^{2\alpha-4} + 1\right) = 2$$

$$\frac{9^{2\alpha-4} + 13}{\frac{5}{2} \cdot 3^{2\alpha-4} + 1} = 4$$

Let $$t = 3^{2\alpha - 4}$$, so $$9^{2\alpha-4} = t^2$$:

$$t^2 + 13 = 4\left(\frac{5t}{2} + 1\right) = 10t + 4$$

$$t^2 - 10t + 9 = 0 \implies (t-1)(t-9) = 0$$

$$t = 1 \Rightarrow 3^{2\alpha-4} = 3^0 \Rightarrow \alpha = 2$$

$$t = 9 \Rightarrow 3^{2\alpha-4} = 3^2 \Rightarrow \alpha = 3$$

So $$S = \{2, 3\}$$.

$$\sum_{\alpha \in S}\alpha = 2 + 3 = 5, \quad \left(\sum_{\alpha \in S}\alpha\right)^2 = 25$$

$$\sum_{\alpha \in S}(\alpha + 1)^2 = 3^2 + 4^2 = 9 + 16 = 25$$

The equation becomes:

$$x^2 - 2(25)x + 25\beta = 0 \implies x^2 - 50x + 25\beta = 0$$

Discriminant $$\geq 0$$:

$$2500 - 100\beta \geq 0 \implies \beta \leq 25$$

The maximum value of $$\beta$$ is $$25$$.