Problem Statement
Find the first four terms of the binomial expansion of $(1 - 2x)^{-1/2}$ in ascending powers of $x$. State the range of values of $x$ for which the expansion is valid.
Verified Solution & Marking Scheme
State generalized binomial series formula
$(1 + u)^p = 1 + pu + \frac{p(p - 1)}{2!}u^2 + \frac{p(p - 1)(p - 2)}{3!}u^3 + \dots$
Here $p = -\frac{1}{2}$ and $u = -2x$.
Compute coefficients term by term
Term 1: $1$
Term 2: $p u = \left(-\frac{1}{2}\right)(-2x) = x$
Term 3: $\frac{(-1/2)(-3/2)}{2} (-2x)^2 = \frac{3/4}{2} (4x^2) = \frac{3}{8}(4x^2) = \frac{3}{2}x^2$
Term 4: $\frac{(-1/2)(-3/2)(-5/2)}{6} (-2x)^3 = \frac{-15/8}{6} (-8x^3) = \left(-\frac{5}{16}\right)(-8x^3) = \frac{5}{2}x^3$
Combine terms and state validity condition
$(1 - 2x)^{-1/2} = 1 + x + \frac{3}{2}x^2 + \frac{5}{2}x^3 + \dots$
Validity requirement: $|u| < 1 \implies |-2x| < 1 \implies 2|x| < 1 \implies |x| < \frac{1}{2}$, or $-\frac{1}{2} < x < \frac{1}{2}$.