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CIE 9231 2026 June Paper 14 Q1

A Level / CIE / FP1

CIE 9231 2026 June Paper 14 Paper · Question 1

题目

Problem

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

∑r=1n(r−1)2=an3+bn2+cn,\sum_{r=1}^{n}(r-1)^2=an^3+bn^2+cn,

where aa, bb and cc are constants to be determined.

(2)

It is given that

ur=1r2−2r+1−1r2+2r+1.u_r=\frac{1}{r^2-2r+1}-\frac{1}{r^2+2r+1}.

(b) Use the method of differences to find ∑r=2nur\displaystyle\sum_{r=2}^{n}u_r in terms of nn.

(4)

(c) State the value of ∑r=2∞ur\displaystyle\sum_{r=2}^{\infty}u_r.

(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)2=an3+bn2+cn,\sum_{r=1}^{n}(r-1)^2=an^3+bn^2+cn,

其中 aa、bb 和 cc 是待求的常数。

(2)

已知

ur=1r2−2r+1−1r2+2r+1.u_r=\frac{1}{r^2-2r+1}-\frac{1}{r^2+2r+1}.

(b) 利用差分法求 ∑r=2nur\displaystyle\sum_{r=2}^{n}u_r，用 nn 表示。

(4)

(c) 写出 ∑r=2∞ur\displaystyle\sum_{r=2}^{\infty}u_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

展开

先展开平方，再分别使用平方和、自然数和与常数项求和公式。

答题过程

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Using the standard results,

∑r=1n(r−1)2= ∑r=1nr2−2∑r=1nr+n= n(n+1)(2n+1)6−n(n+1)+n= 2n3+3n2+n6−n2= 13n3−12n2+16n.\begin{align*} \sum_{r=1}^n(r-1)^2 =&\,\sum_{r=1}^n r^2-2\sum_{r=1}^n r+n \\[4mm] =&\,\frac{n(n+1)(2n+1)}6-n(n+1)+n \\[4mm] =&\,\frac{2n^3+3n^2+n}{6}-n^2 \\[4mm] =&\,\frac13n^3-\frac12n^2+\frac16n. \end{align*}

Hence a=1/3, b=−1/2, c=1/6\boxed{a=1/3,\ b=-1/2,\ c=1/6}.

(b)

解法一

思路

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两个分母都是完全平方。相隔两项的倒数平方会相消，因此首尾各有两项留下。

答题过程

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We have

ur=1(r−1)2−1(r+1)2.u_r=\frac1{(r-1)^2}-\frac1{(r+1)^2}.

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. Using the method of differences,

∑r=2nur= (1−132) +(122−142) +(132−152)+⋯ +(1(n−3)2−1(n−1)2) +(1(n−2)2−1n2) +(1(n−1)2−1(n+1)2).\begin{align*} \sum_{r=2}^n u_r =&\,\bigg(1-\frac1{3^2}\bigg) \\[4mm] &\,\hspace{2pt}+\bigg(\frac1{2^2}-\frac1{4^2}\bigg)\\[4mm] &\,\hspace{4pt}+\bigg(\frac1{3^2}-\frac1{5^2}\bigg)+\cdots \\[4mm] &\,\hspace{2pt}+\bigg(\frac1{(n-3)^2}-\frac1{(n-1)^2}\bigg) \\[4mm] &\,\hspace{4pt}+\bigg(\frac1{(n-2)^2}-\frac1{n^2}\bigg) \\[4mm] &\,\hspace{6pt}+\bigg(\frac1{(n-1)^2}-\frac1{(n+1)^2}\bigg). \end{align*}

Every interior negative term cancels a positive term two positions later. The surviving terms give

∑r=2nur=54−1n2−1(n+1)2.\boxed{\sum_{r=2}^n u_r=\frac54-\frac1{n^2}-\frac1{(n+1)^2}}.

(c)

解法一

思路

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令有限和中的n趋向无穷，两项倒数平方都趋于零。

答题过程

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Taking n→∞n\to\infty in (b),

∑r=2∞ur=54.\boxed{\sum_{r=2}^{\infty}u_r=\frac54}.