利用 $\ln(1+x)$ 和 $\ln(1-x)$ 的标准级数，求出 $\displaystyle\ln\left(\frac{1+x}{1-x}\right)$ 的级数的前三个非零项。 通过在第 (i) 部分的答案中选择合适的 $x$ 值，求 $\displaystyle\sum_{r=0}^{\infty}\frac{1}{(2r+1)4^r}$ 的值。 已知 $\displaystyle\sum_{r=1}^n r^2=\frac{n(n+1)(2n+1)}{6}$，用 $a$ 表示以下式子的值： $\displaystyle\sum_{r=37}^a(2r)^2\,,\quad\text{其中 $a$ 是整数且 $a\ge 37$。}$

1. State the standard series expansions for $\ln(1+x)$ and $\ln(1-x)$ up to the $x^5$ term to ensure three non-zero terms remain after subtraction.

$\ln(1+x) = x - \frac{x^2}{2} + \frac{x^3}{3} - \frac{x^4}{4} + \frac{x^5}{5} - \dots$

Substituting $-x$ for $x$ gives:
$\ln(1-x) = (-x) - \frac{(-x)^2}{2} + \frac{(-x)^3}{3} - \frac{(-x)^4}{4} + \frac{(-x)^5}{5} - \dots$
$\ln(1-x) = -x - \frac{x^2}{2} - \frac{x^3}{3} - \frac{x^4}{4} - \frac{x^5}{5} - \dots$

2. Use logarithm properties to split the expression and substitute the expansions.

$\ln\left(\frac{1+x}{1-x}\right) = \ln(1+x) - \ln(1-x)$

Substituting the series:
$= \left( x - \frac{x^2}{2} + \frac{x^3}{3} - \frac{x^4}{4} + \frac{x^5}{5} - \dots \right) - \left( -x - \frac{x^2}{2} - \frac{x^3}{3} - \frac{x^4}{4} - \frac{x^5}{5} - \dots \right)$

3. Simplify the expression by collecting like terms to find the first three non-zero terms.

$= \left(x - (-x)\right) + \left(-\frac{x^2}{2} - \left(-\frac{x^2}{2}\right)\right) + \left(\frac{x^3}{3} - \left(-\frac{x^3}{3}\right)\right) + \dots$
$= 2x + \frac{2x^3}{3} + \frac{2x^5}{5} + \dots$

1. Concisely obtain the series expansion from part (i) and express it in general summation notation.

From part (i), using standard Maclaurin series:
$\ln\left(\frac{1+x}{1-x}\right) = \ln(1+x) - \ln(1-x)$
$= \left(x - \frac{x^2}{2} + \frac{x^3}{3} - \dots\right) - \left(-x - \frac{x^2}{2} - \frac{x^3}{3} - \dots\right)$
$= 2x + \frac{2x^3}{3} + \frac{2x^5}{5} + \dots$

In sigma notation, this general series is $\displaystyle\sum_{r=0}^{\infty} \frac{2x^{2r+1}}{2r+1}$.

2. Compare the general term of the expansion to the given infinite sum to deduce the suitable value of $x$.

We can factor the series from part (i) to match the target form:
$\displaystyle\sum_{r=0}^{\infty} \frac{2x^{2r+1}}{2r+1} = 2x \sum_{r=0}^{\infty} \frac{x^{2r}}{2r+1} = 2x \sum_{r=0}^{\infty} \frac{1}{2r+1}\left(x^2\right)^r$

We compare this to the target sum:
$\displaystyle\sum_{r=0}^{\infty} \frac{1}{(2r+1)4^r} = \sum_{r=0}^{\infty} \frac{1}{2r+1}\left(\frac{1}{4}\right)^r$

By equating the variable parts:
$x^2 = \frac{1}{4} \implies x = \pm\frac{1}{2}$

To ensure the leading multiplier matches the implied $1$ in the target sum ($2x = 1$), we must choose $x = \frac{1}{2}$.

3. Substitute $x = \frac{1}{2}$ into the logarithmic expression to find the exact value of the sum.

Substitute $x = \frac{1}{2}$ into the original exact logarithmic function:
$\displaystyle\sum_{r=0}^{\infty}\frac{1}{(2r+1)4^r} = \ln\left(\frac{1 + \frac{1}{2}}{1 - \frac{1}{2}}\right)$
$= \ln\left(\frac{3/2}{1/2}\right)$
$= \ln 3$

1. Factor out the constant from the summation and split the summation range.

$\sum_{r=37}^a (2r)^2 = \sum_{r=37}^a 4r^2 = 4 \sum_{r=37}^a r^2$
$= 4 \left( \sum_{r=1}^a r^2 - \sum_{r=1}^{36} r^2 \right)$

2. Apply the given sum formula to both parts.

Using the given formula $\sum_{r=1}^n r^2 = \frac{n(n+1)(2n+1)}{6}$:
$= 4 \left( \frac{a(a+1)(2a+1)}{6} - \frac{36(36+1)(2(36)+1)}{6} \right)$

3. Simplify the expression and compute the constant term.

$= 4 \left( \frac{a(a+1)(2a+1)}{6} - \frac{36(37)(73)}{6} \right)$
$= 4 \left( \frac{a(a+1)(2a+1)}{6} - 16206 \right)$
$= \frac{2}{3}a(a+1)(2a+1) - 64824$