If $$\alpha, \beta$$ are natural numbers such that $$100^\alpha - 199\beta = (100)(100) + (99)(101) + (98)(102) + \ldots + (1)(199)$$, then the slope of the line passing through $$(\alpha, \beta)$$ and origin is:

We need to evaluate the right-hand side: $$(100)(100) + (99)(101) + (98)(102) + \ldots + (1)(199)$$.

The general term is $$(100 - r)(100 + r) = 100^2 - r^2$$ where $$r$$ ranges from 0 to 99. So the sum equals $$\sum_{r=0}^{99}(100^2 - r^2) = 100 \times 100^2 - \sum_{r=0}^{99} r^2$$.

Using the formula $$\sum_{r=0}^{99} r^2 = \sum_{r=1}^{99} r^2 = \frac{99 \times 100 \times 199}{6} = \frac{1970100}{6} = 328350$$.

So the sum is $$1000000 - 328350 = 671650$$.

We are given $$100^{\alpha} - 199\beta = 671650$$. Since $$\alpha$$ and $$\beta$$ are natural numbers, try $$\alpha = 3$$: then $$100^3 - 199\beta = 671650$$, giving $$199\beta = 1000000 - 671650 = 328350$$, so $$\beta = \frac{328350}{199} = 1650$$.

The slope of the line through $$(\alpha, \beta) = (3, 1650)$$ and the origin is $$\frac{\beta}{\alpha} = \frac{1650}{3} = 550$$.

The answer is Option B: $$550$$.