To expand \((1 + 4x)^{-\frac{1}{2}}\), we use the binomial series expansion for \((1 + u)^n\), where \(u = 4x\) and \(n = -\frac{1}{2}\).

The binomial series expansion is:

\((1 + u)^n = 1 + nu + \frac{n(n-1)}{2!}u^2 + \frac{n(n-1)(n-2)}{3!}u^3 + \cdots\)

Substitute \(n = -\frac{1}{2}\) and \(u = 4x\):

\(1 + \left(-\frac{1}{2}\right)(4x) + \frac{-\frac{1}{2}(-\frac{3}{2})}{2!}(4x)^2 + \frac{-\frac{1}{2}(-\frac{3}{2})(-\frac{5}{2})}{3!}(4x)^3\)

Calculate each term:

First term: \(1\)

Second term: \(-2x\)

Third term: \(\frac{(-\frac{1}{2})(-\frac{3}{2})}{2}(16x^2) = 6x^2\)

Fourth term: \(\frac{(-\frac{1}{2})(-\frac{3}{2})(-\frac{5}{2})}{6}(64x^3) = -20x^3\)

Thus, the expansion up to \(x^3\) is:

\(1 - 2x + 6x^2 - 20x^3\)