题目

Problem

(a) Express

\begin{align*} \frac{1}{(3n-1)(3n+5)} \end{align*}

in partial fractions.

(2)

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

\begin{align*} \sum_{r=1}^{n}\frac{20}{(3r-1)(3r+5)} = \frac{n(An+B)}{(3n+C)(3n+D)} \end{align*}

where $A$ , $B$ , $C$ and $D$ are integers to be determined.

(4)

解答

展开

两个一次因式相乘，所以设成两个简单分式。代入方便的 $n$ 值可以快速求出常数。

展开

Let

\begin{align*} \frac{1}{(3n-1)(3n+5)} = \frac{A}{3n-1}+\frac{B}{3n+5} \end{align*}

Then

\begin{align*} 1 ={}&A(3n+5)+B(3n-1) \end{align*}

When $3n-1=0$ , $n=\frac13$ , so

\begin{align*} 1=6A \quad\Longrightarrow\quad A=\frac16 \end{align*}

When $3n+5=0$ , $n=-\frac53$ , so

\begin{align*} 1=-6B \quad\Longrightarrow\quad B=-\frac16 \end{align*}

Therefore

\begin{align*} \frac{1}{(3n-1)(3n+5)} = \frac{1}{6(3n-1)} - \frac{1}{6(3n+5)} \end{align*}

展开

把 (a) 的结果乘以 $20$ 后求和。由于分母相差 $6$ ，展开时中间项会隔项抵消，最后只剩开头两个正项和结尾两个负项。

展开

Using part (a),

\begin{align*} \frac{20}{(3r-1)(3r+5)} ={}& \frac{20}{6} \left( \frac{1}{3r-1} - \frac{1}{3r+5} \right) \end{align*}

Hence

\begin{align*} \sum_{r=1}^{n} \frac{20}{(3r-1)(3r+5)} ={}& \frac{20}{6} \sum_{r=1}^{n} \left( \frac{1}{3r-1} - \frac{1}{3r+5} \right) \end{align*}

Expanding the sum,

\begin{align*} {}& \frac{20}{6} \bigg[ \left(\frac12-\frac18\right) + \left(\frac15-\frac{1}{11}\right)\\[3mm] &\,\hspace{2pt}+ \left(\frac18-\frac{1}{14}\right) +\cdots\\[3mm] &\,\hspace{4pt}+ \left(\frac{1}{3n-7}-\frac{1}{3n-1}\right)\\[3mm] &\,\hspace{6pt}+ \left(\frac{1}{3n-4}-\frac{1}{3n+2}\right)\\[3mm] &\,\hspace{8pt}+ \left(\frac{1}{3n-1}-\frac{1}{3n+5}\right) \bigg] \end{align*}

So all the middle terms cancel and

\begin{align*} \sum_{r=1}^{n} \frac{20}{(3r-1)(3r+5)} ={}& \frac{20}{6} \left( \frac12+\frac15 - \frac{1}{3n+2} - \frac{1}{3n+5} \right)\\[3mm] ={}& \frac{20}{6} \left( \frac{7}{10} - \frac{6n+7}{(3n+2)(3n+5)} \right)\\[3mm] ={}& \frac{20}{6} \left( \frac{7(3n+2)(3n+5)-10(6n+7)} {10(3n+2)(3n+5)} \right)\\[3mm] ={}& \frac{20}{6} \left( \frac{63n^2+87n}{10(3n+2)(3n+5)} \right)\\[3mm] ={}& \frac{n(21n+29)}{(3n+2)(3n+5)} \end{align*}

Therefore

\begin{align*} A=21,\qquad B=29,\qquad C=2,\qquad D=5 \end{align*}