利用公式表（MF26）中的标准级数，将 $\ln(k + 2x)^n$ 按 $x$ 的升幂展开，直到 $x^3$ 项，其中 $k$ 和 $n$ 为实常数，$k > 0$，$n < 0$ 且 $x > -\frac{k}{2}$。 写出第 (i) 部分中展开式成立的 $x$ 的取值范围。 求 $\int_{0}^{1} \ln(1 + 2x)^{-2} \; \mathrm{d}x$ 的精确值。 利用第 (i) 部分中的级数来估计 $\int_{0}^{1} \ln(1 + 2x)^{-2} \; \mathrm{d}x$。 计算第 (iv) 部分中估计值的百分误差，并结合第 (ii) 部分的结论评述该近似值的准确度。

1. Use logarithm properties to rewrite the expression into a form suitable for the standard series expansion.

$\ln(k + 2x)^n = n\ln(k + 2x)$
$= n\ln\left[k\left(1 + \frac{2x}{k}\right)\right]$
$= n\ln k + n\ln\left(1 + \frac{2x}{k}\right)$

2. Apply the standard Maclaurin series expansion for $\ln(1+x)$ from the MF26 formula list.

From MF26, the standard series is $\ln(1+y) = y - \frac{y^2}{2} + \frac{y^3}{3} - \dots$
Substitute $y = \frac{2x}{k}$:
$\ln\left(1 + \frac{2x}{k}\right) = \left(\frac{2x}{k}\right) - \frac{1}{2}\left(\frac{2x}{k}\right)^2 + \frac{1}{3}\left(\frac{2x}{k}\right)^3 - \dots$
$= \frac{2x}{k} - \frac{2x^2}{k^2} + \frac{8x^3}{3k^3} - \dots$

3. Multiply by $n$ and simplify to obtain the final expansion up to the term in $x^3$.

$n\ln(k + 2x) = n\ln k + n\left( \frac{2x}{k} - \frac{2x^2}{k^2} + \frac{8x^3}{3k^3} - \dots \right)$
$= n\ln k + \frac{2n}{k}x - \frac{2n}{k^2}x^2 + \frac{8n}{3k^3}x^3 + \dots$

1. State the condition for the expansion of the logarithmic term to be valid

The expansion in part (i) relies on the standard series for $\ln(1 + y)$ where $y = \frac{2x}{k}$.
The standard expansion for $\ln(1 + y)$ is valid for $-1 < y \le 1$.
Therefore, we require:
$-1 < \frac{2x}{k} \le 1$

2. Solve the inequality for $x$

Since it is given that $k > 0$, we can safely multiply the inequality by $\frac{k}{2}$ without flipping the inequality signs:
$-\frac{k}{2} < x \le \frac{k}{2}$

1. Simplify the integrand using logarithm properties and set up integration by parts.

$\int_{0}^{1} \ln(1 + 2x)^{-2} \; \mathrm{d}x = -2 \int_{0}^{1} \ln(1 + 2x) \; \mathrm{d}x$
Let $u = \ln(1+2x)$ and $\frac{\mathrm{d}v}{\mathrm{d}x} = 1$:
$\frac{\mathrm{d}u}{\mathrm{d}x} = \frac{2}{1+2x}, \quad v = x$

2. Apply the integration by parts formula to the definite integral.

Using $\int u \frac{\mathrm{d}v}{\mathrm{d}x} \mathrm{d}x = uv - \int v \frac{\mathrm{d}u}{\mathrm{d}x} \mathrm{d}x$:
$-2 \int_{0}^{1} \ln(1 + 2x) \; \mathrm{d}x = -2 \left[ x \ln(1+2x) \right]_0^1 - (-2) \int_{0}^{1} x \left(\frac{2}{1+2x}\right) \; \mathrm{d}x$
$= -2(1 \cdot \ln 3 - 0) + \int_{0}^{1} \frac{4x}{1+2x} \; \mathrm{d}x$
$= -2\ln 3 + \int_{0}^{1} \frac{4x}{1+2x} \; \mathrm{d}x$

3. Use algebraic manipulation to integrate the rational fraction.

$\frac{4x}{1+2x} = \frac{2(1+2x) - 2}{1+2x} = 2 - \frac{2}{1+2x}$
$\int_{0}^{1} \left( 2 - \frac{2}{1+2x} \right) \; \mathrm{d}x = \left[ 2x - \ln(1+2x) \right]_0^1$
$= (2 - \ln 3) - (0 - \ln 1)$
$= 2 - \ln 3$

4. Combine the results to obtain the final exact value.

$\text{Exact value} = -2\ln 3 + 2 - \ln 3 = 2 - 3\ln 3$

1. Derive the series expansion for $\ln(k+2x)^n$ required from part (i).

$\text{As required from (i):}$
$\ln(k + 2x)^n = n \ln\left[k\left(1 + \frac{2x}{k}\right)\right] = n \ln k + n\ln\left(1 + \frac{2x}{k}\right)$
$\approx n \ln k + n \left( \frac{2x}{k} - \frac{1}{2}\left(\frac{2x}{k}\right)^2 + \frac{1}{3}\left(\frac{2x}{k}\right)^3 \right)$
$= n \ln k + \frac{2n}{k}x - \frac{2n}{k^2}x^2 + \frac{8n}{3k^3}x^3$

2. Substitute $k=1$ and $n=-2$ into the series expansion to approximate $\ln(1+2x)^{-2}$.

$\text{Matching } \ln(1 + 2x)^{-2} \text{ with } \ln(k + 2x)^n, \text{ let } k = 1 \text{ and } n = -2:$
$\ln(1 + 2x)^{-2} \approx -2 \ln 1 + \frac{2(-2)}{1}x - \frac{2(-2)}{1^2}x^2 + \frac{8(-2)}{3(1^3)}x^3$
$= -4x + 4x^2 - \frac{16}{3}x^3$

3. Evaluate the definite integral of the polynomial approximation from $x=0$ to $x=1$.

$\int_{0}^{1} \ln(1 + 2x)^{-2} \; \mathrm{d}x \approx \int_{0}^{1} \left( -4x + 4x^2 - \frac{16}{3}x^3 \right) \; \mathrm{d}x$
$= \left[ -2x^2 + \frac{4}{3}x^3 - \frac{4}{3}x^4 \right]_{0}^{1}$
$= \left(-2(1)^2 + \frac{4}{3}(1)^3 - \frac{4}{3}(1)^4\right) - (0)$
$= -2$

1. Calculate the exact value and estimated value from parts (iii) and (iv) to be used for the percentage error.

From part (iii), the exact value is:
$\int_{0}^{1} \ln(1 + 2x)^{-2} \; \mathrm{d}x = -2\int_{0}^{1} \ln(1 + 2x) \; \mathrm{d}x$
Using the substitution $u = 1+2x$, $\mathrm{d}u = 2\mathrm{d}x$:
$-\int_{1}^{3} \ln u \; \mathrm{d}u = -[u\ln u - u]_1^3 = 2 - 3\ln 3 \approx -1.2958$

From part (iv), using the series expansion for $k=1, n=-2$:
$\ln(1+2x)^{-2} \approx -4x + 4x^2 - \frac{16}{3}x^3$
$\int_{0}^{1} \left(-4x + 4x^2 - \frac{16}{3}x^3\right) \; \mathrm{d}x = \left[-2x^2 + \frac{4}{3}x^3 - \frac{4}{3}x^4\right]_0^1 = -2$

2. Calculate the percentage error of the estimate.

$\text{Percentage error} = \left| \frac{\text{Approximate} - \text{Exact}}{\text{Exact}} \right| \times 100\%$
$= \left| \frac{-2 - (2 - 3\ln 3)}{2 - 3\ln 3} \right| \times 100\%$
$= \left| \frac{3\ln 3 - 4}{2 - 3\ln 3} \right| \times 100\%$
$\approx \left| \frac{-0.70416}{-1.29584} \right| \times 100\% \approx 54.3\%$

3. Comment on the accuracy using the range of validity from part (ii).

In part (ii), the expansion of $\ln(k+2x)^n = n\ln k + n\ln\left(1+\frac{2x}{k}\right)$ is valid for $-1 < \frac{2x}{k} \le 1 \implies -\frac{k}{2} < x \le \frac{k}{2}$.

For $k=1$, the series is valid only for $-\frac{1}{2} < x \le \frac{1}{2}$.
Since the integration evaluates from $x=0$ to $x=1$, a portion of the interval $[0, 1]$ lies outside the range of validity for the series. Therefore, the series diverges or is grossly inaccurate for $x > 0.5$, resulting in a poor overall approximation (as evidenced by the large $54.3\%$ error).