Choice A is the correct answer. Evaluate at $t = 0$.

1. Substitute: $V(0) = \frac{1,000,000}{1 + 999e^{0}}$.
2. Simplify: $e^0 = 1$, so $V(0) = \frac{1,000,000}{1 + 999} = \frac{1,000,000}{1000}$.
3. Result: $V(0) = 1,000$ views.

💡 Strategic Tip: Logistic functions start at $\frac{K}{1+A}$ when $t=0$.

Choice B is incorrect because this is the carrying capacity (long-term limit). Choice C is incorrect because this is close but not the exact initial value. Choice D is incorrect because this is half the carrying capacity.