题目

Problem

(a) Use standard results from the list of formulae (MF19) to find $\displaystyle\sum_{r=1}^{n}(8r^3+12r^2+4r+3)$ in terms of $n$ , simplifying your answer.

(3)

(b) Show that

\frac{1}{r^4}-\frac{1}{(r+1)^4}=\frac{4r^3+6r^2+4r+1}{r^4(r+1)^4}

and hence use the method of differences to find $\displaystyle\sum_{r=1}^{n}\frac{4r^3+6r^2+4r+1}{r^4(r+1)^4}$ .

(5)

(1)

题目中文翻译

(a) 利用公式表（MF19）中的标准结果，求 $\displaystyle\sum_{r=1}^{n}(8r^3+12r^2+4r+3)$ ，用 $n$ 表示，并化简答案。

(3)

(b) 证明

\frac{1}{r^4}-\frac{1}{(r+1)^4}=\frac{4r^3+6r^2+4r+1}{r^4(r+1)^4},

并由此利用差分法求 $\displaystyle\sum_{r=1}^{n}\frac{4r^3+6r^2+4r+1}{r^4(r+1)^4}$ 。

(5)

(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

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将四项分别求和，套用立方和、平方和及自然数和，再整理同类项。

答题过程

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Using the standard sums,

\begin{align*} \sum_{r=1}^n(8r^3+12r^2+4r+3) =&\,2n^2(n+1)^2+2n(n+1)(2n+1) \\[4mm] &\,\hspace{2pt}+2n(n+1)+3n \\[4mm] =&\,2n^4+4n^3+2n^2+4n^3+6n^2+2n \\[4mm] &\,\hspace{2pt}+2n^2+2n+3n \\[4mm] =&\,\boxed{2n^4+8n^3+10n^2+7n}. \end{align*}

(b)

解法一

思路

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先通分并展开四次方证明恒等式，再用首尾各三项展示相邻项相消。

答题过程

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Putting over a common denominator,

\begin{align*} \frac1{r^4}-\frac1{(r+1)^4} =&\,\frac{(r+1)^4-r^4}{r^4(r+1)^4} \\[4mm] =&\,\frac{4r^3+6r^2+4r+1}{r^4(r+1)^4}, \end{align*}

as required. For $n\ge6$ , the first and last three terms display the cancellation below. For $1\le n\le5$ , directly cancelling the corresponding terms gives the same result. Hence, by the method of differences,

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

All interior fractions cancel in adjacent terms.

(c)

解法一

思路

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承接(b)，有限和的剩余项随n增大趋于零。

答题过程

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Letting $n\to\infty$ in (b),

\boxed{\sum_{r=1}^{\infty}\frac{4r^3+6r^2+4r+1}{r^4(r+1)^4}=1}.

CIE 9231 2025 November 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