Review question

# Can we expand $(1+x)^{3/2} + (1-2x)^{3/4}$ as far as the term in $x^3$? Add to your resource collection Remove from your resource collection Add notes to this resource View your notes for this resource

Ref: R7436

## Solution

Use the binomial theorem to expand $\begin{equation*} (1+x)^\frac{3}{2} + (1-2x)^\frac{3}{4} \end{equation*}$

as far as the term in $x^3$.

The binomial theorem says that, if $|x| < 1$, then $\begin{equation*} (1+x)^r = 1 + rx + \frac{r(r-1)}{2!} x^2 + \frac{r(r-1)(r-2)}{3!} x^3 + \dotsb \end{equation*}$ and so to expand $\begin{equation*} (1+x)^\frac{3}{2} + (1-2x)^\frac{3}{4} \end{equation*}$

as far as the term in $x^3$, we can

1. separately expand $(1+x)^\frac{3}{2}$ and $(1-2x)^\frac{3}{4}$ up to $x^3$ and
2. collect the terms.
Thus \begin{align*} (1+x)^\frac{3}{2} &= 1 + \frac{3}{2}x + \frac{\frac{3}{2} \times \frac{1}{2}}{2!} x^2 + \frac{\frac{3}{2} \times \frac{1}{2} \times \left(- \frac{1}{2} \right)}{3!} x^3 + \ldots\\ &= 1 + \frac{3}{2} x + \frac{3}{8} x^2 - \frac{3}{48} x^3 + \ldots, \end{align*} and \begin{align*} (1-2x)^\frac{3}{4} &= 1 + \frac{3}{4} (-2x) + \frac{\frac{3}{4} \times \left( -\frac{1}{4} \right)}{2!} (-2x)^2 + \frac{\frac{3}{4} \times \left( -\frac{1}{4} \right) \times \left( -\frac{5}{4} \right)}{3!} (-2x)^3 + \ldots\\ &= 1 - \frac{3}{2} x - \frac{3}{8} x^2 - \frac{15}{48} x^3 + \ldots, \end{align*} and therefore \begin{align*} (1+x)^\frac{3}{2} + (1-2x)^\frac{3}{4} &= \left( 1 + \frac{3}{2} x + \frac{3}{8} x^2 - \frac{3}{48} x^3 \right) + \left( 1 - \frac{3}{2} x - \frac{3}{8} x^2 - \frac{15}{48} x^3 \right)+ \ldots \\ &= 2 - \frac{18}{48} x^3 + \ldots \\ &= 2 - \frac{3}{8} x^3 + \ldots \end{align*}