The coefficient of $$x^{101}$$ in the expression
$$5 + x^{500} + x(5 + x)^{499} + x^2(5 + x)^{498} + \ldots + x^{500}, x > 0$$ is

We write the given expression as a sum: $$S = (5 + x)^{500} + x(5 + x)^{499} + x^2(5 + x)^{498} + \ldots + x^{500}$$ which can be expressed as $$S = \sum_{k=0}^{500} x^k (5 + x)^{500-k}\,.$$

Since this is a geometric series with first term $$(5+x)^{500}$$, common ratio $$\frac{x}{5+x}$$, and 501 terms, we have $$S = (5+x)^{500} \cdot \frac{1 - \left(\frac{x}{5+x}\right)^{501}}{1 - \frac{x}{5+x}}\,. $$ Substituting and simplifying gives $$= (5+x)^{500} \cdot \frac{\frac{(5+x)^{501} - x^{501}}{(5+x)^{501}}}{\frac{5}{5+x}} = \frac{(5+x)^{501} - x^{501}}{5}\,. $$

To find the coefficient of $$x^{101}$$, note that the term $$x^{501}$$ contributes zero since 101 < 501. The coefficient of $$x^{101}$$ in $$(5+x)^{501}$$ is $$\binom{501}{101} \cdot 5^{400}\,. $$ Dividing by 5 therefore yields $$\frac{\binom{501}{101} \cdot 5^{400}}{5} = \binom{501}{101} \cdot 5^{399}\,. $$ From the above, the answer is $$^{501}C_{101} \times 5^{399},$$ which is Option A.