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CIE 9231 2025 November Paper 14 Q3

A Level / CIE / FP1

CIE 9231 2025 Nov Paper 14 Paper · Question 3

题目

Problem

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

∑r=1n(r+1)(r+2)(r+3)=14n(n3+bn2+cn+d),\sum_{r=1}^{n}(r+1)(r+2)(r+3)=\frac14n(n^3+bn^2+cn+d),

where bb, cc and dd are integers to be determined.

(3)

(b) Express 2(r+1)(r+2)(r+3)\dfrac{2}{(r+1)(r+2)(r+3)} in partial fractions and hence use the method of differences to find

∑r=1n2(r+1)(r+2)(r+3).\sum_{r=1}^{n}\frac{2}{(r+1)(r+2)(r+3)}.
(5)

(c) State the value of ∑r=1∞2(r+1)(r+2)(r+3)\displaystyle\sum_{r=1}^{\infty}\frac{2}{(r+1)(r+2)(r+3)}.

(1)

题目中文翻译

(a) 利用公式表（MF19）中的标准结果，证明

∑r=1n(r+1)(r+2)(r+3)=14n(n3+bn2+cn+d),\sum_{r=1}^{n}(r+1)(r+2)(r+3)=\frac14n(n^3+bn^2+cn+d),

其中 bb、cc 和 dd 是待求的整数。

(3)

(b) 将 2(r+1)(r+2)(r+3)\dfrac{2}{(r+1)(r+2)(r+3)} 分解为部分分式，并由此利用差分法求

∑r=1n2(r+1)(r+2)(r+3).\sum_{r=1}^{n}\frac{2}{(r+1)(r+2)(r+3)}.
(5)

(c) 写出 ∑r=1∞2(r+1)(r+2)(r+3)\displaystyle\sum_{r=1}^{\infty}\frac{2}{(r+1)(r+2)(r+3)} 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

展开

先展开三次乘积，再套用三个标准和公式，最后整理成题目指定的四分之一n乘多项式。

答题过程

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Expanding,

(r+1)(r+2)(r+3)=r3+6r2+11r+6.(r+1)(r+2)(r+3)=r^3+6r^2+11r+6.

Therefore

∑r=1n(r+1)(r+2)(r+3)= 14n2(n+1)2+n(n+1)(2n+1) +112n(n+1)+6n= 14n(n(n+1)2+4(n+1)(2n+1) +22(n+1)+24)= 14n(n3+10n2+35n+50).\begin{align*} \sum_{r=1}^n(r+1)(r+2)(r+3) =&\,\frac14n^2(n+1)^2+n(n+1)(2n+1) \\[4mm] &\,\hspace{2pt}+\frac{11}{2}n(n+1)+6n \\[4mm] =&\,\frac14n\big(n(n+1)^2+4(n+1)(2n+1) \\[4mm] &\,\hspace{2pt}+22(n+1)+24\big) \\[4mm] =&\,\frac14n(n^3+10n^2+35n+50). \end{align*}

Hence b=10, c=35, d=50\boxed{b=10,\ c=35,\ d=50}.

(b)

解法一

思路

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分式分解后相邻三项的系数是1、−2、1，内部项合计相消，保留首尾四项。

答题过程

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Write

2(r+1)(r+2)(r+3)=Ar+1+Br+2+Cr+3.\frac2{(r+1)(r+2)(r+3)} =\frac A{r+1}+\frac B{r+2}+\frac C{r+3}.

Multiplying by the common denominator and setting r=−1,−2,−3r=-1,-2,-3 gives A=1A=1, B=−2B=-2, C=1C=1. Hence

2(r+1)(r+2)(r+3)=1r+1−2r+2+1r+3.\frac2{(r+1)(r+2)(r+3)} =\frac1{r+1}-\frac2{r+2}+\frac1{r+3}.

For n≥6n\ge6, the first and last three terms display the cancellation below. For 1≤n≤51\le n\le5, directly cancelling the corresponding terms gives the same result. By the method of differences,

∑r=1n2(r+1)(r+2)(r+3)= (12−23+14) +(13−24+15) +(14−25+16)+⋯ +(1n−1−2n+1n+1) +(1n−2n+1+1n+2) +(1n+1−2n+2+1n+3).\begin{align*} \sum_{r=1}^n\frac2{(r+1)(r+2)(r+3)} =&\,\bigg(\frac12-\frac23+\frac14\bigg) \\[4mm] &\,\hspace{2pt}+\bigg(\frac13-\frac24+\frac15\bigg) \\[4mm] &\,\hspace{2pt}+\bigg(\frac14-\frac25+\frac16\bigg)+\cdots \\[4mm] &\,\hspace{4pt}+\bigg(\frac1{n-1}-\frac2n+\frac1{n+1}\bigg) \\[4mm] &\,\hspace{6pt}+\bigg(\frac1n-\frac2{n+1}+\frac1{n+2}\bigg) \\[4mm] &\,\hspace{8pt}+\bigg(\frac1{n+1}-\frac2{n+2}+\frac1{n+3}\bigg). \end{align*}

The interior coefficients cancel since 1−2+1=01-2+1=0. The surviving terms give

∑r=1n2(r+1)(r+2)(r+3)= 12−13−1n+2+1n+3= 16−1n+2+1n+3.\begin{align*} \sum_{r=1}^n\frac2{(r+1)(r+2)(r+3)} =&\,\frac12-\frac13-\frac1{n+2}+\frac1{n+3}\\[4mm] =&\,\boxed{\frac16-\frac1{n+2}+\frac1{n+3}}. \end{align*}

(c)

解法一

思路

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在有限和中令n趋向无穷即可。

答题过程

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Taking n→∞n\to\infty in (b),

∑r=1∞2(r+1)(r+2)(r+3)=16.\boxed{\sum_{r=1}^{\infty}\frac2{(r+1)(r+2)(r+3)}=\frac16}.