题目

Problem

(a) Show that, for $r > 0$

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

(2)

(b) Hence show that

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

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

(4)

解答

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这是一个 show that，不能只说“显然”。把左边通分到共同分母 $r(r+1)(r+2)$ ，自然得到右边。

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Using the common denominator $r(r+1)(r+2)$ ,

\begin{align*} \frac{r+2}{r(r+1)}-\frac{r+3}{(r+1)(r+2)} =&\,\frac{(r+2)^2-r(r+3)}{r(r+1)(r+2)}\\[4mm] =&\,\frac{r^2+4r+4-r^2-3r}{r(r+1)(r+2)}\\[4mm] =&\,\frac{r+4}{r(r+1)(r+2)}. \end{align*}

This proves the required identity.

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这里的 Hence 表示必须使用 (a) 的拆分。把前几项和最后几项写出来，就能看见中间项相消，只剩第一项和最后一项。望远镜级数最怕省略号写得太快，所以这里要把首几项和末几项都展示出来。

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From part (a),

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

Hence,

\begin{align*} &\,\sum_{r=1}^{n}\frac{r+4}{r(r+1)(r+2)}\\[4mm] =&\,\left(\frac{3}{1\cdot2}-\frac{4}{2\cdot3}\right)\\[4mm] &\,+\left(\frac{4}{2\cdot3}-\frac{5}{3\cdot4}\right)\\[4mm] &\,+\left(\frac{5}{3\cdot4}-\frac{6}{4\cdot5}\right)\\[4mm] &\,+\cdots\\[4mm] &\,+\left(\frac{n+1}{(n-1)n}\right.\\[4mm] &\,\hspace{18pt}\left.-\frac{n+2}{n(n+1)}\right)\\[4mm] &\,+\left(\frac{n+2}{n(n+1)}\right.\\[4mm] &\,\hspace{18pt}\left.-\frac{n+3}{(n+1)(n+2)}\right)\\[4mm] =&\,\frac{3}{2}-\frac{n+3}{(n+1)(n+2)}. \end{align*}

Putting this over a common denominator,

\begin{align*} &\,\frac{3}{2}-\frac{n+3}{(n+1)(n+2)}\\[4mm] =&\,\frac{3(n+1)(n+2)-2(n+3)}{2(n+1)(n+2)}\\[4mm] =&\,\frac{3(n^2+3n+2)-2n-6}{2(n+1)(n+2)}\\[4mm] =&\,\frac{3n^2+9n+6-2n-6}{2(n+1)(n+2)}\\[4mm] =&\,\frac{3n^2+7n}{2(n+1)(n+2)}\\[4mm] =&\,\frac{n(3n+7)}{2(n+1)(n+2)}. \end{align*}

Therefore this is in the required form

\begin{align*} \frac{n(an+b)}{c(n+1)(n+2)} \end{align*}

with

\begin{align*} a=3,\qquad b=7,\qquad c=2. \end{align*}