题目

Problem

Given that

\begin{align*} \frac{2n + 1}{n^2 (n + 1)^2} \equiv \frac{A}{n^2} + \frac{B}{(n + 1)^2} \end{align*}

(a) determine the value of $A$ and the value of $B$

(1)

(b) Hence show that, for $n \geqslant 5$

\begin{align*} \sum_{r=5}^n \frac{2r + 1}{r^2 (r + 1)^2} = \frac{n^2 + an + b}{c(n + 1)^2} \end{align*}

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

(4)

解答

展开

右边通分后比较分子即可。因为分母已经相同，所以只需要比较

\begin{align*} 2n+1 \end{align*}

和展开后的分子。

展开

Start with

\begin{align*} \frac{2n+1}{n^2(n+1)^2} \equiv \frac{A}{n^2}+\frac{B}{(n+1)^2}. \end{align*}

Use the common denominator $n^2(n+1)^2$ :

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

\begin{align*} 2n+1 \equiv&\,A(n+1)^2+Bn^2\\[2mm] =&\,A(n^2+2n+1)+Bn^2\\[2mm] =&\,(A+B)n^2+2An+A. \end{align*}

Compare coefficients:

\begin{align*} A=1,\qquad 2A=2,\qquad A+B=0. \end{align*}

Therefore

\begin{align*} A=1,\qquad B=-1. \end{align*}

展开

由 (a) 可得

\begin{align*} \frac{2r+1}{r^2(r+1)^2} =\frac1{r^2}-\frac1{(r+1)^2}. \end{align*}

这是望远镜求和。把 $r=5$ 到 $n$ 展开后，中间项会全部抵消。

展开

From part (a),

\begin{align*} \frac{2r+1}{r^2(r+1)^2} =&\,\frac1{r^2}-\frac1{(r+1)^2}. \end{align*}

\begin{align*} &\,\sum_{r=5}^{n} \frac{2r+1}{r^2(r+1)^2}\\[4mm] =&\,\sum_{r=5}^{n} \left(\frac1{r^2}-\frac1{(r+1)^2}\right). \end{align*}

Write out the terms:

\begin{align*} &\,\left(\frac1{5^2}-\frac1{6^2}\right)\\[4mm] &\,\hspace{2pt}+\left(\frac1{6^2}-\frac1{7^2}\right)\\[4mm] &\,\hspace{4pt}+\cdots\\[4mm] &\,\hspace{6pt}+\left(\frac1{n^2}-\frac1{(n+1)^2}\right). \end{align*}

The middle terms cancel, so

\begin{align*} \sum_{r=5}^{n} \frac{2r+1}{r^2(r+1)^2} =&\,\frac1{25}-\frac1{(n+1)^2}. \end{align*}

Put over a common denominator:

\begin{align*} \frac1{25}-\frac1{(n+1)^2} =&\,\frac{(n+1)^2-25}{25(n+1)^2}\\[2mm] =&\,\frac{n^2+2n+1-25}{25(n+1)^2}\\[2mm] =&\,\frac{n^2+2n-24}{25(n+1)^2}. \end{align*}

Thus

\begin{align*} a=2,\qquad b=-24,\qquad c=25. \end{align*}