设 $a$ 为正常数。

(a) 用裂项相消法求 $\displaystyle\sum_{r=1}^{n}\frac{1}{(ar+1)(ar+a+1)}$，用 $n$ 和 $a$ 表示。

(b) 求使 $\displaystyle\sum_{r=1}^{\infty}\frac{1}{(ar+1)(ar+a+1)}=\frac16$ 成立的 $a$ 的值。

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。

相邻两个一次因式相差 $a$，先裂项，再把开头和结尾的项展开，展示中间项如何相消。

Since $a>0$,

Thus, for $n\ge6$,

The intermediate terms cancel in adjacent pairs. For small $n$, the displayed beginning and end overlap; each summand is included only once. The same cancellation, or direct substitution, confirms the result for $n=1,2,3,4,5$.

沿用有限和，先利用 $a>0$ 取极限，再解二次方程并排除非正根。

As $n\to\infty$, $1/(an+a+1)\to0$ because $a>0$. Hence

Therefore

Since $a>0$, $\boxed{a=2}$.