已知 $\dfrac{1}{a+bx}$ 展开式的前三项为 $\dfrac{1}{2}+\dfrac{1}{4}x+cx^2$，求 $a$、$b$ 和 $c$ 的值。 利用第 (i) 小题的答案，求由曲线 $y = \dfrac{1}{a+bx}$、两条坐标轴以及直线 $x = \tfrac{1}{2}$ 所围成区域面积的近似值。以 $\dfrac{p}{q}$ 的形式给出你的答案，其中 $p$ 和 $q$ 为整数。由此求出 $\ln \dfrac{4}{3}$ 的一个估计值。

1. Rewrite the expression and apply the binomial expansion up to the $x^2$ term.

2. Compare the constant term to find the value of $a$.

Comparing the constant term with $\frac{1}{2} + \frac{1}{4}x + cx^2$:
$\frac{1}{a} = \frac{1}{2} \implies a = 2$

3. Compare the coefficient of $x$ to find the value of $b$.

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

4. Compare the coefficient of $x^2$ to find the value of $c$.

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

1. Solve for $a$, $b$, and $c$ from part (i) as prerequisites for part (ii).

$(a+bx)^{-1} = a^{-1}\left(1 + \frac{b}{a}x\right)^{-1} = \frac{1}{a} - \frac{b}{a^2}x + \frac{b^2}{a^3}x^2 - \dots$
Comparing with $\frac{1}{2} + \frac{1}{4}x + cx^2$:
$\frac{1}{a} = \frac{1}{2} \implies a = 2$
$-\frac{b}{a^2} = \frac{1}{4} \implies -\frac{b}{4} = \frac{1}{4} \implies b = -1$
$c = \frac{b^2}{a^3} = \frac{(-1)^2}{2^3} = \frac{1}{8}$

2. Set up and evaluate the definite integral of the series expansion to find an approximate area.

$\text{Approximate Area} = \int_0^{\frac{1}{2}} \left( \frac{1}{2} + \frac{1}{4}x + \frac{1}{8}x^2 \right) \text{d}x$
$= \left[ \frac{1}{2}x + \frac{1}{8}x^2 + \frac{1}{24}x^3 \right]_0^{\frac{1}{2}}$
$= \frac{1}{2}\left(\frac{1}{2}\right) + \frac{1}{8}\left(\frac{1}{4}\right) + \frac{1}{24}\left(\frac{1}{8}\right)$
$= \frac{1}{4} + \frac{1}{32} + \frac{1}{192}$
$= \frac{48 + 6 + 1}{192} = \frac{55}{192}$

3. Evaluate the exact definite integral of the curve to find the exact area.

$\text{Exact Area} = \int_0^{\frac{1}{2}} \frac{1}{2-x} \text{d}x$
$= \left[ -\ln|2-x| \right]_0^{\frac{1}{2}}$
$= -\ln\left(2 - \frac{1}{2}\right) - (-\ln 2)$
$= -\ln\left(\frac{3}{2}\right) + \ln 2 = \ln\left(\frac{2}{\frac{3}{2}}\right) = \ln \frac{4}{3}$

4. Equate the exact area to the approximate area to find an estimate for $\ln \frac{4}{3}$.

$\text{Exact Area} \approx \text{Approximate Area}$
$\ln \frac{4}{3} \approx \frac{55}{192}$