Let $$f : (0, 2) \to R$$ be defined as $$f(x) = \log_2\left(1 + \tan\left(\frac{\pi x}{4}\right)\right)$$. Then, $$\lim_{n \to \infty} \frac{2}{n}\left(f\left(\frac{1}{n}\right) + f\left(\frac{2}{n}\right) + \ldots + f(1)\right)$$ is equal to ________.

We have $$f(x) = \log_2\left(1 + \tan\left(\frac{\pi x}{4}\right)\right)$$ and need to evaluate $$\lim_{n \to \infty} \frac{2}{n}\left(f\left(\frac{1}{n}\right) + f\left(\frac{2}{n}\right) + \ldots + f(1)\right)$$.

This expression is a Riemann sum. Writing $$\frac{2}{n}\sum_{k=1}^{n} f\left(\frac{k}{n}\right) = 2 \cdot \frac{1}{n}\sum_{k=1}^{n} f\left(\frac{k}{n}\right)$$, which converges to $$2\int_0^1 f(x)\,dx$$ as $$n \to \infty$$.

So we need $$I = 2\int_0^1 \log_2\left(1 + \tan\left(\frac{\pi x}{4}\right)\right) dx$$.

Let $$u = \frac{\pi x}{4}$$, so $$du = \frac{\pi}{4}dx$$, and when $$x = 0$$, $$u = 0$$; when $$x = 1$$, $$u = \frac{\pi}{4}$$.

$$I = 2 \cdot \frac{4}{\pi} \int_0^{\pi/4} \log_2(1 + \tan u)\,du = \frac{8}{\pi} \int_0^{\pi/4} \log_2(1 + \tan u)\,du$$.

We use the well-known result: $$\int_0^{\pi/4} \ln(1 + \tan u)\,du = \frac{\pi}{8}\ln 2$$. This follows from the substitution $$u \to \frac{\pi}{4} - u$$: $$1 + \tan\left(\frac{\pi}{4} - u\right) = 1 + \frac{1 - \tan u}{1 + \tan u} = \frac{2}{1 + \tan u}$$, so $$\int_0^{\pi/4}\ln(1+\tan u)\,du = \int_0^{\pi/4}\ln 2\,du - \int_0^{\pi/4}\ln(1+\tan u)\,du$$, giving $$2\int_0^{\pi/4}\ln(1+\tan u)\,du = \frac{\pi}{4}\ln 2$$.

Since $$\log_2(1 + \tan u) = \frac{\ln(1 + \tan u)}{\ln 2}$$, we get $$\int_0^{\pi/4} \log_2(1 + \tan u)\,du = \frac{1}{\ln 2} \cdot \frac{\pi \ln 2}{8} = \frac{\pi}{8}$$.

Therefore $$I = \frac{8}{\pi} \cdot \frac{\pi}{8} = 1$$.