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CIE 9231 2023 November Paper 12 Q1

A Level / CIE / FP1

CIE 9231 2023 Nov Paper 12 Paper · Question 1

题目

Problem

(a) Use standard results from the list of formulae (MF19) to find ∑r=1n(3r2+3r+1)\displaystyle\sum_{r=1}^{n}(3r^2+3r+1) in terms of nn, simplifying your answer.

(3)

(b) Show that

1r3−1(r+1)3=3r2+3r+1r3(r+1)3\frac{1}{r^3}-\frac{1}{(r+1)^3}=\frac{3r^2+3r+1}{r^3(r+1)^3}

and hence use the method of differences to find ∑r=1n3r2+3r+1r3(r+1)3\displaystyle\sum_{r=1}^{n}\frac{3r^2+3r+1}{r^3(r+1)^3}.

(5)

(c) Deduce the value of ∑r=1∞3r2+3r+1r3(r+1)3\displaystyle\sum_{r=1}^{\infty}\frac{3r^2+3r+1}{r^3(r+1)^3}.

(1)
题目中文翻译

(a) 利用公式表（MF19）中的标准结果，用 nn 表示 ∑r=1n(3r2+3r+1)\displaystyle\sum_{r=1}^{n}(3r^2+3r+1)，并化简你的答案。

(3)

(b) 证明

1r3−1(r+1)3=3r2+3r+1r3(r+1)3\frac{1}{r^3}-\frac{1}{(r+1)^3}=\frac{3r^2+3r+1}{r^3(r+1)^3}

并由此利用差分法求 ∑r=1n3r2+3r+1r3(r+1)3\displaystyle\sum_{r=1}^{n}\frac{3r^2+3r+1}{r^3(r+1)^3}。

(5)

(c) 推导 ∑r=1∞3r2+3r+1r3(r+1)3\displaystyle\sum_{r=1}^{\infty}\frac{3r^2+3r+1}{r^3(r+1)^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

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分别代入平方和、自然数和以及常数求和公式，再提取公因式化简。

答题过程

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Using the standard sum formulae,

∑r=1n(3r2+3r+1)= 12n(n+1)(2n+1)+32n(n+1)+n= 12n(n+1)(2n+4)+n= n(n+1)(n+2)+n= n3+3n2+3n.\begin{align*} \sum_{r=1}^n(3r^2+3r+1) =&\,\frac12n(n+1)(2n+1)+\frac32n(n+1)+n\\[4mm] =&\,\frac12n(n+1)(2n+4)+n\\[4mm] =&\,n(n+1)(n+2)+n\\[4mm] =&\,\boxed{n^3+3n^2+3n}. \end{align*}

(b)

解法一

思路

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通分并展开立方差证明恒等式，再逐项写出相消。

答题过程

展开 1r3−1(r+1)3= (r+1)3−r3r3(r+1)3= 3r2+3r+1r3(r+1)3,\begin{align*} \frac1{r^3}-\frac1{(r+1)^3} =&\,\frac{(r+1)^3-r^3}{r^3(r+1)^3}\\[4mm] =&\,\frac{3r^2+3r+1}{r^3(r+1)^3}, \end{align*}

as required. Hence

Sn= (1−123) +(123−133) +(133−143)+⋯ +(1(n−2)3−1(n−1)3) +(1(n−1)3−1n3) +(1n3−1(n+1)3)= 1−1(n+1)3.\begin{align*} S_n=&\,\bigg(1-\frac1{2^3}\bigg)\\[4mm] &\,\hspace{2pt}+\bigg(\frac1{2^3}-\frac1{3^3}\bigg)\\[4mm] &\,\hspace{4pt}+\bigg(\frac1{3^3}-\frac1{4^3}\bigg)+\cdots\\[4mm] &\,\hspace{2pt}+\bigg(\frac1{(n-2)^3}-\frac1{(n-1)^3}\bigg)\\[4mm] &\,\hspace{4pt}+\bigg(\frac1{(n-1)^3}-\frac1{n^3}\bigg)\\[4mm] &\,\hspace{6pt}+\bigg(\frac1{n^3}-\frac1{(n+1)^3}\bigg)\\[4mm] =&\,\boxed{1-\frac1{(n+1)^3}}. \end{align*}

The expansion illustrates cancellation for sufficiently large nn; the resulting formula also holds directly for smaller positive nn.

(c)

解法一

思路

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承接部分和，取n趋于无穷，余项倒数趋于零。

答题过程

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From (b), the infinite series has sum

lim⁡n→∞Sn=1−lim⁡n→∞1(n+1)3=1.\boxed{\lim_{n\to\infty}S_n =1-\lim_{n\to\infty}\frac1{(n+1)^3}=1}.