Given that the first three terms in the expansion of $\dfrac{1}{a+bx}$ are $\dfrac{1}{2}+\dfrac{1}{4}x+cx^2$, find the values of $a$, $b$ and $c$. Use your answer in part (i) to find an approximation to the area of the region bounded by the curve $y=\dfrac{1}{a+bx}$, the two axes and the line $x=\dfrac{1}{2}$. Give your answer in the form $\dfrac{p}{q}$ where $p$ and $q$ are integers. Hence find an estimate for $\ln\dfrac{4}{3}$.

1. Factor out $a$ to apply the standard binomial expansion for $(1+x)^n$.

$\begin{aligned}\frac{1}{a+bx} &= (a+bx)^{-1} \\ &= a^{-1}\left(1+\frac{b}{a}x\right)^{-1} \\ &= \frac{1}{a}\left( 1 + (-1)\left(\frac{b}{a}x\right) + \frac{(-1)(-2)}{2!}\left(\frac{b}{a}x\right)^2 + \dots \right) \\ &= \frac{1}{a} - \frac{b}{a^2}x + \frac{b^2}{a^3}x^2 + \dots\end{aligned}$

2. Compare the constant term and the $x$ coefficient to find $a$ and $b$.

Comparing the constant term with $\frac{1}{2}$:
$\frac{1}{a} = \frac{1}{2} \implies a = 2$

Comparing the coefficient of $x$ with $\frac{1}{4}$:
$-\frac{b}{a^2} = \frac{1}{4}$
Substitute $a=2$:
$-\frac{b}{2^2} = \frac{1}{4} \implies -b = 1 \implies b = -1$

3. Compare the $x^2$ coefficient to evaluate $c$.

Comparing the coefficient of $x^2$ with $c$:
$c = \frac{b^2}{a^3}$
Substitute $a=2$ and $b=-1$:
$c = \frac{(-1)^2}{2^3} = \frac{1}{8}$

1. Retrieve the values of $a$, $b$, and $c$ found in part (i).

From part (i), expanding $\frac{1}{a+bx} = a^{-1}\left(1 + \frac{b}{a}x\right)^{-1} \approx \frac{1}{a} - \frac{b}{a^2}x + \frac{b^2}{a^3}x^2$.
Comparing coefficients with $\frac{1}{2}+\frac{1}{4}x+cx^2$ yields $a=2$, $b=-1$, and $c=\frac{1}{8}$.

2. Integrate the polynomial expansion to find an approximation for the area under the curve.

The area bounded by the curve, the $x$-axis, the $y$-axis ($x=0$), and the line $x=\frac{1}{2}$ is approximated by integrating the expansion from $0$ to $\frac{1}{2}$:
$\text{Area} \approx \int_0^{\frac{1}{2}} \left(\frac{1}{2} + \frac{1}{4}x + \frac{1}{8}x^2\right) \, dx$
$= \left[ \frac{1}{2}x + \frac{1}{8}x^2 + \frac{1}{24}x^3 \right]_0^{\frac{1}{2}}$
$= \left( \frac{1}{2}\left(\frac{1}{2}\right) + \frac{1}{8}\left(\frac{1}{4}\right) + \frac{1}{24}\left(\frac{1}{8}\right) \right) - 0$
$= \frac{1}{4} + \frac{1}{32} + \frac{1}{192}$
$= \frac{48}{192} + \frac{6}{192} + \frac{1}{192} = \frac{55}{192}$

3. Find the exact area by integrating the original function to reveal the logarithmic term.

The exact area is given by the integral of the original function $y = \frac{1}{2-x}$:
$\text{Area} = \int_0^{\frac{1}{2}} \frac{1}{2-x} \, dx$
$= \left[ -\ln(2-x) \right]_0^{\frac{1}{2}}$
$= -\ln\left(2 - \frac{1}{2}\right) - (-\ln 2)$
$= -\ln\left(\frac{3}{2}\right) + \ln 2$
$= \ln\left(\frac{2}{\frac{3}{2}}\right) = \ln\left(\frac{4}{3}\right)$

4. Equate the exact exact area expression to the approximated area to estimate $\ln\frac{4}{3}$.

Since the truncated series approximates the original function, their integrals over the same limits are approximately equal:
$\ln\left(\frac{4}{3}\right) \approx \frac{55}{192}$