The coefficient of $$x^{301}$$ in $$1 + x^{500} + x \cdot 1 + x^{499} + x^2 \cdot 1 + x^{498} + \ldots + x^{500}$$ is:

Each subsequent term is formed by multiplying the previous term by

$$\frac{x}{1+x}$$

Hence, the given series is a geometric progression (G.P.).

First term:

$$a = (1+x)^{500}$$

Common ratio:

$$r = \frac{x}{1+x}$$

Number of terms:

$$n = 501$$

The sum of a finite G.P. is

$$S_n = a\left(\frac{1-r^n}{1-r}\right)$$

Substituting the values:

$$S = (1+x)^{500}\left(\frac{1-\left(\frac{x}{1+x}\right)^{501}}{1-\frac{x}{1+x}}\right)$$

Simplifying the denominator:

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

Therefore,

$$S = \frac{(1+x)^{500}\left[1-\frac{x^{501}}{(1+x)^{501}}\right]}{\frac{1}{1+x}}$$

Dividing by a fraction means multiplying by its reciprocal:

$$S = (1+x)^{500}(1+x)\left[1-\frac{x^{501}}{(1+x)^{501}}\right]$$

$$S = (1+x)^{501}\left[1-\frac{x^{501}}{(1+x)^{501}}\right]$$

Distributing:

$$S = (1+x)^{501} - (1+x)^{501}\cdot\frac{x^{501}}{(1+x)^{501}}$$

$$S = (1+x)^{501} - x^{501}$$

We need the coefficient of $$x^{301}$$.

The term $$-x^{501}$$ does not affect the coefficient of $$x^{301}$$.

Hence, we only need the coefficient of $$x^{301}$$ in

$$(1+x)^{501}$$

Using the Binomial Theorem:

$$(1+x)^n = \sum_{k=0}^{n}\binom{n}{k}x^k$$

Therefore, the coefficient of $$x^{301}$$ is

$$\binom{501}{301}$$

Using symmetry of binomial coefficients,

$$\binom{501}{301} = \binom{501}{200}$$

Hence, the required option is D.