题目

Problem

(a) Express

\begin{align*} \frac{4r+2}{r(r+1)(r+2)} \end{align*}

in partial fractions.

(3)

(b) Hence, using the method of differences, prove that

\begin{align*} \sum_{r=1}^{n} \frac{4r+2}{r(r+1)(r+2)} = \frac{n(an+b)}{2(n+1)(n+2)} \end{align*}

where $a$ and $b$ are constants to be found.

(5)

解答

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分母是三个连续的一次因式，所以设成三项简单分式。代入 $r=0,-1,-2$ 可以很快求出三个常数。

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Let

\begin{align*} \frac{4r+2}{r(r+1)(r+2)} = \frac{A}{r}+\frac{B}{r+1}+\frac{C}{r+2} \end{align*}

Then

\begin{align*} 4r+2 = A(r+1)(r+2)+Br(r+2)+Cr(r+1) \end{align*}

Put $r=0$ :

\begin{align*} 2=2A \quad\Longrightarrow\quad A=1 \end{align*}

Put $r=-1$ :

\begin{align*} -2=-B \quad\Longrightarrow\quad B=2 \end{align*}

Put $r=-2$ :

\begin{align*} -6=2C \quad\Longrightarrow\quad C=-3 \end{align*}

Therefore

\begin{align*} \frac{4r+2}{r(r+1)(r+2)} = \frac1r+\frac{2}{r+1}-\frac{3}{r+2} \end{align*}

展开

把 (a) 的结果逐项展开。中间大部分项会抵消，剩下开头的常数项和结尾含 $n$ 的项，再通分整理。

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Using part (a),

\begin{align*} \sum_{r=1}^{n} \frac{4r+2}{r(r+1)(r+2)} = \sum_{r=1}^{n} \left( \frac1r+\frac{2}{r+1}-\frac{3}{r+2} \right) \end{align*}

Expanding,

\begin{align*} \sum_{r=1}^{n} \left( \frac1r+\frac{2}{r+1}-\frac{3}{r+2} \right) ={}& \left(1+\frac22-\frac33\right) + \left(\frac12+\frac23-\frac34\right) + \left(\frac13+\frac24-\frac35\right) +\cdots\\[3mm] &\,\hspace{2pt}+ \left(\frac{1}{n-1}+\frac{2}{n}-\frac{3}{n+1}\right) + \left(\frac{1}{n}+\frac{2}{n+1}-\frac{3}{n+2}\right) \end{align*}

After cancellation, only the first three positive terms and the last three negative terms remain:

\begin{align*} \sum_{r=1}^{n} \frac{4r+2}{r(r+1)(r+2)} ={}& 1+\frac22+\frac12 -\frac{3}{n+1} +\frac{2}{n+1} -\frac{3}{n+2}\\[3mm] ={}& \frac{5}{2}-\frac{1}{n+1}-\frac{3}{n+2} \end{align*}

Put over a common denominator:

\begin{align*} \frac52-\frac{1}{n+1}-\frac{3}{n+2} ={}& \frac{5(n+1)(n+2)-2(n+2)-6(n+1)} {2(n+1)(n+2)}\\[3mm] ={}& \frac{5n^2+15n+10-2n-4-6n-6} {2(n+1)(n+2)}\\[3mm] ={}& \frac{5n^2+7n}{2(n+1)(n+2)}\\[3mm] ={}& \frac{n(5n+7)}{2(n+1)(n+2)} \end{align*}

Therefore

\begin{align*} a=5,\qquad b=7 \end{align*}