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CIE 9231 2025 June Paper 14 Q1

A Level / CIE / FP1

CIE 9231 2025 June Paper 14 Paper · Question 1

题目

Problem

(a) Use the List of formulae (MF19) to find ∑r=1n(2r+1)\displaystyle\sum_{r=1}^{n}(2r+1) in terms of nn, simplifying your answer.

(2)

(b) Show that

2r+1(r2+1)(r2+2r+2)=1r2+1−1r2+2r+2.\frac{2r+1}{(r^2+1)(r^2+2r+2)}=\frac{1}{r^2+1}-\frac{1}{r^2+2r+2}.
(1)

(c) Use the method of differences to find ∑r=1n2r+1(r2+1)(r2+2r+2)\displaystyle\sum_{r=1}^{n}\frac{2r+1}{(r^2+1)(r^2+2r+2)}.

(2)

(d) Deduce the value of ∑r=1∞2r+1(r2+1)(r2+2r+2)\displaystyle\sum_{r=1}^{\infty}\frac{2r+1}{(r^2+1)(r^2+2r+2)}.

(1)
题目中文翻译

(a) 利用公式表（MF19）求 ∑r=1n(2r+1)\displaystyle\sum_{r=1}^{n}(2r+1)，以 nn 表示，并化简答案。

(2)

(b) 证明

2r+1(r2+1)(r2+2r+2)=1r2+1−1r2+2r+2.\frac{2r+1}{(r^2+1)(r^2+2r+2)}=\frac{1}{r^2+1}-\frac{1}{r^2+2r+2}.
(1)

(c) 用差分法求 ∑r=1n2r+1(r2+1)(r2+2r+2)\displaystyle\sum_{r=1}^{n}\frac{2r+1}{(r^2+1)(r^2+2r+2)}。

(2)

(d) 由此求 ∑r=1∞2r+1(r2+1)(r2+2r+2)\displaystyle\sum_{r=1}^{\infty}\frac{2r+1}{(r^2+1)(r^2+2r+2)} 的值。

(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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把求和拆成自然数之和与常数之和，再提取公因式。

答题过程

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Using ∑r=1nr=12n(n+1)\sum_{r=1}^n r=\frac12n(n+1),

∑r=1n(2r+1)= 2∑r=1nr+∑r=1n1= n(n+1)+n= n(n+2).\begin{align*} \sum_{r=1}^n(2r+1)=&\,2\sum_{r=1}^n r+\sum_{r=1}^n1\\[4mm] =&\,n(n+1)+n\\[4mm] =&\,\boxed{n(n+2)}. \end{align*}

(b)

解法一

思路

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右边通分，分子的平方项和常数相消后恰为2r+1。

答题过程

展开 1r2+1−1r2+2r+2= r2+2r+2−(r2+1)(r2+1)(r2+2r+2)= 2r+1(r2+1)(r2+2r+2),\begin{align*} \frac1{r^2+1}-\frac1{r^2+2r+2} =&\,\frac{r^2+2r+2-(r^2+1)}{(r^2+1)(r^2+2r+2)}\\[4mm] =&\,\frac{2r+1}{(r^2+1)(r^2+2r+2)}, \end{align*}

which is the required identity.

(c)

解法一

思路

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第二个分母等于下一个整数的平方加1。因此相邻括号的负项和正项相消，只剩首项与末项。

答题过程

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By (b), the summand is 1r2+1−1(r+1)2+1\frac1{r^2+1}-\frac1{(r+1)^2+1}. Thus

Sn= (12−15) +(15−110) +(110−117)+⋯ +(1(n−2)2+1−1(n−1)2+1) +(1(n−1)2+1−1n2+1) +(1n2+1−1(n+1)2+1)= 12−1n2+2n+2.\begin{align*} S_n=&\,\bigg(\frac12-\frac15\bigg)\\[4mm] &\,\hspace{2pt}+\bigg(\frac15-\frac1{10}\bigg)\\[4mm] &\,\hspace{4pt}+\bigg(\frac1{10}-\frac1{17}\bigg)+\cdots\\[4mm] &\,\hspace{2pt}+\bigg(\frac1{(n-2)^2+1}-\frac1{(n-1)^2+1}\bigg)\\[4mm] &\,\hspace{4pt}+\bigg(\frac1{(n-1)^2+1}-\frac1{n^2+1}\bigg)\\[4mm] &\,\hspace{6pt}+\bigg(\frac1{n^2+1}-\frac1{(n+1)^2+1}\bigg)\\[4mm] =&\,\boxed{\frac12-\frac1{n^2+2n+2}}. \end{align*}

The displayed cancellation applies for sufficiently large nn; the same formula holds directly for the smaller positive integers.

(d)

解法一

思路

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无穷级数的和是部分和的极限，末项分母趋于无穷。

答题过程

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From (c),

lim⁡n→∞Sn=12−lim⁡n→∞1n2+2n+2=12.\lim_{n\to\infty}S_n =\frac12-\lim_{n\to\infty}\frac1{n^2+2n+2} =\boxed{\frac12}.