题目

Problem

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

\begin{align*} &\,\frac{r+2}{r(r+1)}\\[4mm] &\,\hspace{2pt}-\frac{r+3}{(r+1)(r+2)}\\[4mm] =&\,\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)}\\[4mm] =&\,\frac{n(an+b)}{c(n+1)(n+2)} \end{align*}

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

(4)

题目中文翻译

2. (a) 证明：当 $r>0$ 时，

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

(2)

(b) 由此证明

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

其中 $a$ 、 $b$ 、 $c$ 是待求的整数。

(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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Let $D=r(r+1)(r+2)$ . Using this common denominator,

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

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

Then

\begin{align*} S_n=&\,\sum_{r=1}^n F_r\\[4mm] =&\,\Big(\frac3{1\cdot2}-\frac4{2\cdot3}\Big)\\[4mm] &\,+\Big(\frac4{2\cdot3}-\frac5{3\cdot4}\Big)\\[4mm] &\,+\Big(\frac5{3\cdot4}-\frac6{4\cdot5}\Big)\\[4mm] &\,+\cdots\\[4mm] &\,+\Big(\frac{n+1}{(n-1)n}\\[2mm] &\,\hspace{2pt}-\frac{n+2}{n(n+1)}\Big)\\[4mm] &\,+\Big(\frac{n+2}{n(n+1)}\\[2mm] &\,\hspace{2pt}-\frac{n+3}{(n+1)(n+2)}\Big)\\[4mm] =&\,\frac32-\frac{n+3}{(n+1)(n+2)}. \end{align*}

Put this over the common denominator $2(n+1)(n+2)$ . Its numerator is

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

Therefore

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

This is the required form, with

\begin{align*} a=&\,3,\\[2mm] b=&\,7,\\[2mm] c=&\,2. \end{align*}