What is this algebraic fraction if its expansion has no $x$ term?

What is this algebraic fraction if its expansion has no $x$ term?

Question

In the expansion of $\frac{1+3x}{(1+ax)^2}$ in ascending powers of $x$, the coefficient of $x$ is zero. Find the value of the constant $a$, and, when $a$ has this value, obtain the expansion up to and including the term in $x^3$.

Solution

We first need to find an expansion for the denominator using the standard binomial expansion.

The binomial theorem says that, if $|x| < 1$, then $(1+x)^r = 1 + rx + \frac{r(r-1)}{2!} x^2 + \frac{r(r-1)(r-2)}{3!} x^3 + \dots b$

In our case,

$r=-2$

and we have

$ax$

in place of

$x$

, so

$$\begin{align*} (1+ax)^{-2} &= 1 -2ax + \frac{-2(-3)}{2!}(ax)^2 + \frac{-2(-3)(-4)}{3!} (ax)^3 + \dots b \\ &=1-2ax+3a^2x^2-4a^3x^3 +\dots b \end{align*}$$

The term in $x$ of the expansion of $\dfrac{1+3x}{(1+ax)^2}$ comes from $1 \times (-2ax) + 3x \times 1$. If the coefficient is zero then we know that $3-2a=0$ so $a=\dfrac{3}{2}$.

…and, when $a$ has this value, obtain the expansion up to and including the term in $x^3$.

Multiplying by the numerator we get $\frac{1+3x}{(1+ax)^2} = 1 + (3-2a)x+(3a^2-6a)x^2+(9a^2-4a^3)x^3+ \dots b$ and substituting for $a$ we get $\frac{1+3x}{(1+\frac{3}{2}x)^2}=1-\frac{9}{4}x^2+\frac{27}{4}x^3 \quad\text{ up to and including the term in } x^3.$

Note that the condition for the expansion to be valid is $\big|ax\big|<1$ which means that $\big|x\big|<\frac{2}{3}$.