Let's break down and verify the series expansions provided:

1. For $\ln(1 + 4x)$: $\ln(1 + 4x) \approx 4x - \frac{(4x)^2}{2} + O(x^3) = 4x - 8x^2 + O(x^3)$ This expansion is derived using the Taylor series expansion for $\ln(1 + u)$, where $u = 4x$: $\ln(1 + u) = u - \frac{u^2}{2} + \frac{u^3}{3} - \cdots$ Substituting $u = 4x$: $\ln(1 + 4x) = 4x - \frac{(4x)^2}{2} + O(x^3) = 4x - 8x^2 + O(x^3)$

2. For $e^{2x} - 1$: $e^{2x} - 1 \approx 2x + \frac{(2x)^2}{2} + O(x^3) = 2x + 2x^2 + O(x^3)$ This expansion is derived using the Taylor series expansion for $e^u$, where $u = 2x$: $e^u = 1 + u + \frac{u^2}{2!} + \frac{u^3}{3!} + \cdots$ Substituting $u = 2x$: $e^{2x} = 1 + 2x + \frac{(2x)^2}{2!} + O(x^3) = 1 + 2x + 2x^2 + O(x^3)$ Subtracting 1 from both sides: $e^{2x} - 1 = 2x + 2x^2 + O(x^3)$

3. For $(1 + 2x)^3 - 1$: $(1 + 2x)^3 - 1 = 1 + 6x + 12x^2 + 8x^3 - 1 = 6x + 12x^2 + 8x^3$ This expansion is derived using the binomial theorem for $(1 + u)^n$, where $u = 2x$ and $n = 3$: $(1 + u)^n = 1 + \binom{n}{1}u + \binom{n}{2}u^2 + \binom{n}{3}u^3 + \cdots$ Substituting $u = 2x$ and $n = 3$: $(1 + 2x)^3 = 1 + 3 \cdot 2x + 3 \cdot (2x)^2 + (2x)^3 = 1 + 6x + 12x^2 + 8x^3$ Subtracting 1 from both sides: [ (1 + 2x)^3 - 1 = 6x + 12x^2 + 8x^3