题目

Problem

(a) Express

\begin{align*} \frac{1}{(2n - 1)(2n + 1)(2n + 3)} \end{align*}

in partial fractions.

(2)

(b) Hence, using the method of differences, show that for all integer values of $n$ ,

\begin{align*} \sum_{r=1}^{n} \frac{1}{(2r - 1)(2r + 1)(2r + 3)} =\,& \frac{n(n + 2)}{a(2n + b)(2n + c)} \end{align*}

where $a$ , $b$ and $c$ are integers to be determined.

(4)

解答

展开

分母由三个一次因式组成，所以设成三个简单分式。代入使其中一个因式为零的 $n$ 值，可以快速求出系数。

展开

Let

\begin{align*} \frac{1}{(2n-1)(2n+1)(2n+3)} =&\,\frac{A}{2n-1}+\frac{B}{2n+1} +\frac{C}{2n+3}. \end{align*}

Multiplying by $(2n-1)(2n+1)(2n+3)$ ,

\begin{align*} 1=&\,A(2n+1)(2n+3)\\[2mm] &\,\hspace{2pt}+B(2n-1)(2n+3)\\[2mm] &\,\hspace{4pt}+C(2n-1)(2n+1). \end{align*}

Put $n=\frac{1}{2}$ :

\begin{align*} 1=A(2)(4) \quad\Longrightarrow\quad A=\frac{1}{8}. \end{align*}

Put $n=-\frac{1}{2}$ :

\begin{align*} 1=B(-2)(2) \quad\Longrightarrow\quad B=-\frac{1}{4}. \end{align*}

Put $n=-\frac{3}{2}$ :

\begin{align*} 1=C(-4)(-2) \quad\Longrightarrow\quad C=\frac{1}{8}. \end{align*}

Therefore

\begin{align*} \boxed{ \frac{1}{(2n-1)(2n+1)(2n+3)} =\frac{1}{8(2n-1)} -\frac{1}{4(2n+1)} +\frac{1}{8(2n+3)} }. \end{align*}

展开

把 (a) 的结果写成

\begin{align*} \frac{1}{8}\left(\frac{1}{2r-1} -\frac{2}{2r+1} +\frac{1}{2r+3}\right). \end{align*}

然后展开前几项和最后几项，保留下没有抵消的项。

展开

From part (a),

\begin{align*} \frac{1}{(2r-1)(2r+1)(2r+3)} =&\,\frac{1}{8}\left(\frac{1}{2r-1} -\frac{2}{2r+1} +\frac{1}{2r+3}\right). \end{align*}

\begin{align*} &\,\sum_{r=1}^{n} \frac{1}{(2r-1)(2r+1)(2r+3)}\\[2mm] =&\,\frac{1}{8}\sum_{r=1}^{n} \left(\frac{1}{2r-1}-\frac{2}{2r+1} +\frac{1}{2r+3}\right). \end{align*}

Write out the terms inside the bracket:

\begin{align*} &\,\left(1-\frac{2}{3}+\frac{1}{5}\right)\\[4mm] &\,\hspace{2pt}+\left(\frac{1}{3}-\frac{2}{5}+\frac{1}{7}\right)\\[4mm] &\,\hspace{4pt}+\left(\frac{1}{5}-\frac{2}{7}+\frac{1}{9}\right)\\[4mm] &\,\hspace{6pt}+\cdots\\[4mm] &\,\hspace{8pt}+\left(\frac{1}{2n-3} -\frac{2}{2n-1}+\frac{1}{2n+1}\right)\\[4mm] &\,\hspace{10pt}+\left(\frac{1}{2n-1} -\frac{2}{2n+1}+\frac{1}{2n+3}\right). \end{align*}

After cancellation,

\begin{align*} \sum_{r=1}^{n} \frac{1}{(2r-1)(2r+1)(2r+3)} =&\,\frac{1}{8} \left(1-\frac{1}{3} -\frac{1}{2n+1}+\frac{1}{2n+3}\right)\\[2mm] =&\,\frac{1}{8} \left(\frac{2}{3} -\frac{2}{(2n+1)(2n+3)}\right). \end{align*}

Combine into a single fraction:

\begin{align*} \frac{1}{8} \left(\frac{2}{3} -\frac{2}{(2n+1)(2n+3)}\right) =&\,\frac{1}{8} \left( \frac{2(2n+1)(2n+3)-6} {3(2n+1)(2n+3)} \right)\\[2mm] =&\,\frac{1}{8} \left( \frac{2(4n^2+8n+3)-6} {3(2n+1)(2n+3)} \right)\\[2mm] =&\,\frac{1}{8} \left( \frac{8n^2+16n} {3(2n+1)(2n+3)} \right)\\[2mm] =&\,\frac{n(n+2)}{3(2n+1)(2n+3)}. \end{align*}

Thus

\begin{align*} \boxed{a=3,\qquad b=1,\qquad c=3}. \end{align*}