(a) 用差分法求 $\displaystyle\sum_{r=1}^{n}\frac{1}{(kr+1)(kr-k+1)}$，用 $n$ 和 $k$ 表示，其中 $k$ 为正常数。

(b) 推导 $\displaystyle\sum_{r=1}^{\infty}\frac{1}{(kr+1)(kr-k+1)}$ 的值。

(c) 另求 $\displaystyle\sum_{r=n}^{n^2}\frac{1}{(kr+1)(kr-k+1)}$，用 $n$ 和 $k$ 表示。

Answer all questions. You should use a calculator where appropriate.

You must show all necessary working clearly; no marks will be given for unsupported answers from a calculator.

Give non-exact numerical answers correct to 3 significant figures, or 1 decimal place for angles in degrees, unless a different level of accuracy is specified in the question.

请回答所有题目。适当时应使用计算器。必须清楚展示所有必要的计算过程；仅给出计算器答案而没有过程，不予评分。除非题目另有精度要求，非精确数值答案保留 3 位有效数字，以度为单位的角度保留 1 位小数。

两个分母相差k，部分分式拆成相邻倒数之差，常数因子为1/k。

Since

the sum is

The expanded pattern applies for sufficiently large $n$; the resulting formula also holds directly for smaller positive integers. All denominators are positive since $k>0$ and $r\geqslant1$.

k为正常数，随着n增大末项倒数趋于零。

From (a), as $n\to\infty$,

取前n平方项之和减前n−1项之和，注意后一项的分母是k(n−1)+1。

Using (a), with $S_0=0$,