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CIE 9231 2022 November Paper 12 Q1

A Level / CIE / FP1

CIE 9231 2022 Nov Paper 12 Paper · Question 1

题目

Problem

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

(2)

(b) Express 1r(r+2)\dfrac{1}{r(r+2)} in partial fractions and hence find ∑r=1n1r(r+2)\displaystyle\sum_{r=1}^{n}\frac{1}{r(r+2)} in terms of nn.

(4)

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

(1)
题目中文翻译

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

(2)

(b) 将 1r(r+2)\dfrac{1}{r(r+2)} 分解为部分分式,并由此求 ∑r=1n1r(r+2)\displaystyle\sum_{r=1}^{n}\frac{1}{r(r+2)},用 nn 表示。

(4)

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

展开

展开r(r+2),代入平方和与自然数和,提取公共因子。

答题过程

展开 ∑r=1nr(r+2)= ∑r=1nr2+2∑r=1nr= n(n+1)(2n+1)6+n(n+1)= 16n(n+1)(2n+7).\begin{align*} \sum_{r=1}^n r(r+2) =&\,\sum_{r=1}^n r^2+2\sum_{r=1}^n r\\[4mm] =&\,\frac{n(n+1)(2n+1)}6+n(n+1)\\[4mm] =&\,\boxed{\frac16n(n+1)(2n+7)}. \end{align*}

(b)

解法一

思路

展开

两个因子相隔2,拆成倒数差后隔两项相消,最后留下开头两项和末尾两项。

答题过程

展开

Let

1r(r+2)=Ar+Br+2.\frac1{r(r+2)}=\frac A r+\frac B{r+2}.

Then 1=A(r+2)+Br1=A(r+2)+Br, giving A+B=0A+B=0 and 2A=12A=1. Hence

1r(r+2)=12(1r−1r+2).\boxed{\frac1{r(r+2)}=\frac12\bigg(\frac1r-\frac1{r+2}\bigg)}.

Therefore

Sn= 12[(1−13) +(12−14) +(13−15)+⋯ +(1n−2−1n) +(1n−1−1n+1) +(1n−1n+2)]= 12(32−1n+1−1n+2).\begin{align*} S_n=&\,\frac12\bigg[\bigg(1-\frac13\bigg)\\[4mm] &\,\hspace{2pt}+\bigg(\frac12-\frac14\bigg)\\[4mm] &\,\hspace{4pt}+\bigg(\frac13-\frac15\bigg)+\cdots\\[4mm] &\,\hspace{6pt}+\bigg(\frac1{n-2}-\frac1n\bigg)\\[4mm] &\,\hspace{8pt}+\bigg(\frac1{n-1}-\frac1{n+1}\bigg)\\[4mm] &\,\hspace{10pt}+\bigg(\frac1n-\frac1{n+2}\bigg)\bigg]\\[4mm] =&\,\boxed{\frac12\bigg(\frac32-\frac1{n+1}-\frac1{n+2}\bigg)}. \end{align*}

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

(c)

解法一

思路

展开

两个末项倒数都趋于零,故无穷和是剩余首项的一半。

答题过程

展开

Using (b),

lim⁡n→∞Sn=12×32=34.\boxed{\lim_{n\to\infty}S_n=\frac12\times\frac32=\frac34}.