The sum of the series $$\frac{1}{x+1} + \frac{2}{x^2+1} + \frac{2^2}{x^4+1} + \ldots + \frac{2^{100}}{x^{2^{100}}+1}$$ when $$x = 2$$ is:

The given series is:

$$S = \frac{1}{x+1} + \frac{2}{x^2+1} + \frac{2^2}{x^4+1} + \dots + \frac{2^{100}}{x^{2^{100}}+1}$$

Consider the following partial fraction identity:

$$\frac{k}{x^k-1} - \frac{k}{x^k+1} = \frac{k(x^k+1) - k(x^k-1)}{x^{2k}-1} = \frac{2k}{x^{2k}-1}$$

This identity shows that subtracting a term with a positive sign from a term with a negative sign (in the denominator) doubles the constant and squares the variable.

We begin by subtracting the series from the term $$\frac{1}{x-1}$$ :

$$\text{Let } S_n = \frac{1}{x-1} - \left[ \frac{1}{x+1} + \frac{2}{x^2+1} + \frac{4}{x^4+1} + \dots + \frac{2^n}{x^{2^n}+1} \right]$$

Applying the identity to the first two terms:

$$\frac{1}{x-1} - \frac{1}{x+1} = \frac{2}{x^2-1}$$

Now, combine this result with the next term of the series:

$$\frac{2}{x^2-1} - \frac{2}{x^2+1} = \frac{4}{x^4-1}$$

Continuing this process for each term up to$$2^{100}$$, 

we find that the entire bracketed sum, when combined with $$\frac{1}{x-1}$$, reduces to:

$$\frac{1}{x-1} - S = \frac{2^{101}}{x^{2^{101}}-1}$$

Rearranging the equation to isolate the sum $$S$$:

$$S = \frac{1}{x-1} - \frac{2^{101}}{x^{2^{101}}-1}$$

Now, substitute the given value $$x = 2$$ :

$$S = \frac{1}{2-1} - \frac{2^{101}}{2^{2^{101}}-1}$$

$$S = 1 - \frac{2^{101}}{2^{2 \cdot 2^{100}}-1}$$

$$S = 1 - \frac{2^{101}}{(2^2)^{2^{100}}-1}$$

$$S = 1 - \frac{2^{101}}{4^{2^{100}}-1}$$

Final Result

The sum of the series is:

$$1 - \frac{2^{101}}{4^{101}-1}$$

The correct option is A.