Let $$A = \begin{pmatrix} 1 & 2 \\ -2 & -5 \end{pmatrix}$$. Let $$\alpha, \beta \in \mathbb{R}$$ be such that $$\alpha A^2 + \beta A = 2I$$. Then $$\alpha + \beta$$ is equal to

We need to evaluate:

$$L = \lim_{n \to \infty} \frac{1}{2^n}\left(\frac{1}{\sqrt{1 - \frac{1}{2^n}}} + \frac{1}{\sqrt{1 - \frac{2}{2^n}}} + \ldots + \frac{1}{\sqrt{1 - \frac{2^n - 1}{2^n}}}\right)$$

Letting $$N = 2^n$$ transforms this into

$$L = \lim_{N \to \infty} \frac{1}{N}\sum_{k=1}^{N-1} \frac{1}{\sqrt{1 - k/N}},$$

which is the Riemann sum for the integral

$$\int_0^1 \frac{1}{\sqrt{1-x}}\, dx.$$

To evaluate this integral, set $$u = 1 - x$$ so that $$du = -dx$$, giving

$$\int_0^1 \frac{1}{\sqrt{1-x}}\, dx = \int_1^0 \frac{-du}{\sqrt{u}} = \int_0^1 \frac{du}{\sqrt{u}} = \left[2\sqrt{u}\right]_0^1 = 2(1) - 2(0) = 2.$$

Therefore, the answer is Option D: $$\textbf{2}$$.