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CIE 9231 2025 November Paper 12 Q1

A Level / CIE / FP1

CIE 9231 2025 Nov Paper 12 Paper · Question 1

题目

Problem

(a) Use standard results from the list of formulae (MF19) to find ∑r=1n(r3−r)\displaystyle\sum_{r=1}^{n}(r^3-r) in terms of nn, fully factorising your answer.

(3)

(b) Express r+3r3−r\dfrac{r+3}{r^3-r} in the form Ar−1+Br+Cr+1\dfrac{A}{r-1}+\dfrac{B}{r}+\dfrac{C}{r+1}, where AA, BB and CC are constants to be determined, and hence use the method of differences to find ∑r=2nr+3r3−r\displaystyle\sum_{r=2}^{n}\frac{r+3}{r^3-r}.

(5)

(c) Deduce the value of ∑r=2∞r+3r3−r\displaystyle\sum_{r=2}^{\infty}\frac{r+3}{r^3-r}.

(1)

题目中文翻译

(a) 利用公式表（MF19）中的标准结果，求 ∑r=1n(r3−r)\displaystyle\sum_{r=1}^{n}(r^3-r)，用 nn 表示，并将答案完全因式分解。

(3)

(b) 将 r+3r3−r\dfrac{r+3}{r^3-r} 表示为 Ar−1+Br+Cr+1\dfrac{A}{r-1}+\dfrac{B}{r}+\dfrac{C}{r+1} 的形式，其中 AA、BB 和 CC 是待求的常数，并由此利用差分法求 ∑r=2nr+3r3−r\displaystyle\sum_{r=2}^{n}\frac{r+3}{r^3-r}。

(5)

(c) 由此推得 ∑r=2∞r+3r3−r\displaystyle\sum_{r=2}^{\infty}\frac{r+3}{r^3-r} 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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用立方和减自然数和，再把公共因子提出并完全分解。

答题过程

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Using the standard formulae,

∑r=1n(r3−r)= 14n2(n+1)2−12n(n+1)= 14n(n+1)(n(n+1)−2)= 14n(n+1)(n−1)(n+2).\begin{align*} \sum_{r=1}^n(r^3-r) =&\,\frac14n^2(n+1)^2-\frac12n(n+1) \\[4mm] =&\,\frac14n(n+1)\big(n(n+1)-2\big) \\[4mm] =&\,\boxed{\frac14n(n+1)(n-1)(n+2)}. \end{align*}

(b)

解法一

思路

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先按三个线性因子分式分解；求和后中间每项的系数合计为零，首尾留下四项。

答题过程

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Since r3−r=(r−1)r(r+1)r^3-r=(r-1)r(r+1), the partial fractions require

r+3= Ar(r+1)+B(r−1)(r+1) +C(r−1)r.\begin{align*} r+3=&\,Ar(r+1)+B(r-1)(r+1)\\[4mm] &\,+C(r-1)r. \end{align*}

Setting r=1,0,−1r=1,0,-1 gives A=2A=2, B=−3B=-3, C=1C=1. Thus

r+3r3−r=2r−1−3r+1r+1.\boxed{\frac{r+3}{r^3-r}=\frac2{r-1}-\frac3r+\frac1{r+1}}.

For n≥7n\ge7, the first and last three terms display the cancellation below. For 2≤n≤62\le n\le6, directly cancelling the corresponding terms gives the same result. Hence, by the method of differences,

∑r=2nr+3r3−r= (2−32+13) +(22−33+14) +(23−34+15)+⋯ +(2n−3−3n−2+1n−1) +(2n−2−3n−1+1n) +(2n−1−3n+1n+1).\begin{align*} \sum_{r=2}^n\frac{r+3}{r^3-r} =&\,\bigg(2-\frac32+\frac13\bigg) \\[4mm] &\,\hspace{2pt}+\bigg(\frac22-\frac33+\frac14\bigg) \\[4mm] &\,\hspace{2pt}+\bigg(\frac23-\frac34+\frac15\bigg)+\cdots \\[4mm] &\,\hspace{4pt}+\bigg(\frac2{n-3}-\frac3{n-2}+\frac1{n-1}\bigg) \\[4mm] &\,\hspace{6pt}+\bigg(\frac2{n-2}-\frac3{n-1}+\frac1n\bigg) \\[4mm] &\,\hspace{8pt}+\bigg(\frac2{n-1}-\frac3n+\frac1{n+1}\bigg). \end{align*}

The interior coefficients cancel because 2−3+1=02-3+1=0. The remaining terms give

∑r=2nr+3r3−r= 2+−3+22 +1−3n+1n+1= 32−2n+1n+1.\begin{align*} \sum_{r=2}^n\frac{r+3}{r^3-r} =&\,2+\frac{-3+2}{2}\\[4mm] &\,+\frac{1-3}{n}+\frac1{n+1} \\[4mm] =&\,\boxed{\frac32-\frac2n+\frac1{n+1}}. \end{align*}

(c)

解法一

思路

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承接有限和，两个含n的分式都趋于零。

答题过程

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Taking n→∞n\to\infty in (b),

∑r=2∞r+3r3−r=32.\boxed{\sum_{r=2}^{\infty}\frac{r+3}{r^3-r}=\frac32}.