Can we expand $(1+x)^{3/2} + (1-2x)^{3/4}$ as far as the term in $x^3$?

Can we expand $(1+x)^{3/2} + (1-2x)^{3/4}$ as far as the term in $x^3$?

Question

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 $x^3$.

Solution

The

binomial theorem

says that, if

$|x| < 1$

, then

$$\begin{equation*} (1+x)^r = 1 + rx + \frac{r(r-1)}{2!} x^2 + \frac{r(r-1)(r-2)}{3!} x^3 + \dots b \end{equation*}$$

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

---

1. 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

$$\begin{align*} (1+x)^\frac{3}{2} + (1-2x)^\frac{3}{4} &= \left( 1 + \frac{3}{2} x + \frac{3}{8} x^2 - \frac{3}{48} x^3 \right) + \left( 1 - \frac{3}{2} x - \frac{3}{8} x^2 - \frac{15}{48} x^3 \right)+ \ldots \\ &= 2 - \frac{18}{48} x^3 + \ldots \\ &= 2 - \frac{3}{8} x^3 + \ldots \end{align*}$$