题目

Problem

(a) Use standard results from the list of formulae (MF19) to show that

\sum_{r=1}^{n}(4r-3)(12r+3)=an^3+bn^2+cn,

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

(3)

(b) Use the method of differences to find $\displaystyle\sum_{r=1}^{n}\frac{1}{(4r-3)(12r+3)}$ in terms of $n$ .

(4)

(1)

题目中文翻译

(a) 使用公式表（MF19）中的标准结果，证明

\sum_{r=1}^{n}(4r-3)(12r+3)=an^3+bn^2+cn,

其中 $a$ 、 $b$ 和 $c$ 为待求的整数。

(3)

(b) 使用差分法，求 $\displaystyle\sum_{r=1}^{n}\frac{1}{(4r-3)(12r+3)}$ ，用 $n$ 表示。

(4)

(1)

原卷通用作答要求

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。

解答

(a)

解法一

思路

展开

先展开二次乘积，再分别套用平方和、自然数和公式，最后整理成题目指定的三次式。

答题过程

展开

Expanding,

(4r-3)(12r+3)=48r^2-24r-9.

Using the standard formulae,

\begin{align*} \sum_{r=1}^n r=&\,\frac{n(n+1)}2,\\[4mm] \sum_{r=1}^n r^2=&\,\frac{n(n+1)(2n+1)}6, \end{align*}

we obtain, writing $S_n=\sum_{r=1}^n(4r-3)(12r+3)$ ,

\begin{align*} S_n=&\,8n(n+1)(2n+1)\\[4mm] &\,-12n(n+1)-9n \\[4mm] =&\,16n^3+24n^2+8n\\[4mm] &\,-12n^2-12n-9n \\[4mm] =&\,16n^3+12n^2-13n. \end{align*}

Hence $\boxed{a=16,\ b=12,\ c=-13}$ .

(b)

解法一

思路

展开

把12r+3写成3(4r+1)，分式分解后相邻项恰好相消。展开首尾各三项展示消去结构。

答题过程

展开

Since $12r+3=3(4r+1)$ ,

\frac1{(4r-3)(12r+3)} =\frac1{12}\bigg(\frac1{4r-3}-\frac1{4r+1}\bigg).

For $n\ge6$ , the first and last three terms give the following cancellation. For $1\le n\le5$ , directly cancelling the corresponding terms gives the same result. By the method of differences,

\begin{align*} \sum_{r=1}^n\frac1{(4r-3)(12r+3)} =&\,\frac1{12}\bigg[\bigg(1-\frac15\bigg) +\bigg(\frac15-\frac19\bigg) \\[4mm] &\,\hspace{2pt}+\bigg(\frac19-\frac1{13}\bigg)+\cdots \\[4mm] &\,\hspace{4pt}+\bigg(\frac1{4n-11}-\frac1{4n-7}\bigg) \\[4mm] &\,\hspace{6pt}+\bigg(\frac1{4n-7}-\frac1{4n-3}\bigg) \\[4mm] &\,\hspace{8pt}+\bigg(\frac1{4n-3}-\frac1{4n+1}\bigg)\bigg] \\[4mm] =&\,\frac1{12}\bigg(1-\frac1{4n+1}\bigg). \end{align*}

Each negative interior fraction cancels the positive fraction in the next term. Thus

\boxed{\sum_{r=1}^n\frac1{(4r-3)(12r+3)}=\frac{n}{3(4n+1)}}.

(c)

解法一

思路

展开

承接有限和，令n趋向无穷。

答题过程

展开

Taking the limit of the result in (b),

\boxed{\sum_{r=1}^{\infty}\frac1{(4r-3)(12r+3)}=\frac1{12}}.

CIE 9231 2026 June Paper 11 Q1

题目

解答

(a)

解法一

思路

答题过程

(b)

解法一

思路

答题过程

(c)

解法一

思路

答题过程