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CIE 9231 2024 June Paper 11 Q3

A Level / CIE / FP1

CIE 9231 2024 June Paper 11 Paper · Question 3

题目

Problem

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

∑r=1Nr(r+1)(3r+4)=112N(N+1)(N+2)(9N+19).\sum_{r=1}^{N}r(r+1)(3r+4)=\frac1{12}N(N+1)(N+2)(9N+19).
(3)

(b) Express 3r+4r(r+1)\dfrac{3r+4}{r(r+1)} in partial fractions and hence use the method of differences to find

∑r=1N3r+4r(r+1)(14)r+1\sum_{r=1}^{N}\frac{3r+4}{r(r+1)}\bigg(\frac14\bigg)^{r+1}

in terms of NN.

(4)

(c) Deduce the value of ∑r=1∞3r+4r(r+1)(14)r+1\displaystyle\sum_{r=1}^{\infty}\frac{3r+4}{r(r+1)}\bigg(\frac14\bigg)^{r+1}.

(1)
题目中文翻译

(a) 使用公式表（MF19）中的标准结果证明

∑r=1Nr(r+1)(3r+4)=112N(N+1)(N+2)(9N+19).\sum_{r=1}^{N}r(r+1)(3r+4)=\frac1{12}N(N+1)(N+2)(9N+19).
(3)

(b) 将 3r+4r(r+1)\dfrac{3r+4}{r(r+1)} 分解为部分分式，并由此用差分法求

∑r=1N3r+4r(r+1)(14)r+1,\sum_{r=1}^{N}\frac{3r+4}{r(r+1)}\bigg(\frac14\bigg)^{r+1},

以 NN 表示。

(4)

(c) 推导出 ∑r=1∞3r+4r(r+1)(14)r+1\displaystyle\sum_{r=1}^{\infty}\frac{3r+4}{r(r+1)}\bigg(\frac14\bigg)^{r+1} 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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先展开成三次、二次和一次项，再代入三个标准求和公式。提取N(N+1)后，剩余二次式因式分解。

答题过程

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Since r(r+1)(3r+4)=3r3+7r2+4rr(r+1)(3r+4)=3r^3+7r^2+4r, the standard formulae give

∑r=1Nr(r+1)(3r+4)= 3(N(N+1)2)2 +7N(N+1)(2N+1)6 +4N(N+1)2= N(N+1)12[9N(N+1) +14(2N+1)+24]= N(N+1)12(9N2+37N+38)= 112N(N+1)(N+2)(9N+19),\begin{align*} \sum_{r=1}^N r(r+1)(3r+4) =&\,3\bigg(\frac{N(N+1)}2\bigg)^2\\[4mm] &\,\hspace{2pt}+7\frac{N(N+1)(2N+1)}6\\[4mm] &\,\hspace{4pt}+4\frac{N(N+1)}2\\[4mm] =&\,\frac{N(N+1)}{12}\big[9N(N+1)\\[4mm] &\,\hspace{2pt}+14(2N+1)+24\big]\\[4mm] =&\,\frac{N(N+1)}{12}(9N^2+37N+38)\\[4mm] =&\,\boxed{\frac1{12}N(N+1)(N+2)(9N+19)}, \end{align*}

as required.

(b)

解法一

思路

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部分分式的首项系数4恰好抵消幂中的一个四分之一，形成带权相邻差。

答题过程

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Put

3r+4r(r+1)=Ar+Br+1.\frac{3r+4}{r(r+1)}=\frac A r+\frac B{r+1}.

Then 3r+4=A(r+1)+Br3r+4=A(r+1)+Br, so A=4A=4 and A+B=3A+B=3, giving B=−1B=-1. Hence

3r+4r(r+1)=4r−1r+1.\boxed{\frac{3r+4}{r(r+1)}=\frac4r-\frac1{r+1}}.

The weighted summand becomes

(4r−1r+1)14r+1=1r4r−1(r+1)4r+1.\bigg(\frac4r-\frac1{r+1}\bigg)\frac1{4^{r+1}} =\frac1{r4^r}-\frac1{(r+1)4^{r+1}}.

Consequently,

SN= (14−12⋅42) +(12⋅42−13⋅43) +(13⋅43−14⋅44)+⋯ +(1(N−2)4N−2−1(N−1)4N−1) +(1(N−1)4N−1−1N4N) +(1N4N−1(N+1)4N+1)= 14−14N+1(N+1).\begin{align*} S_N=&\,\bigg(\frac14-\frac1{2\cdot4^2}\bigg)\\[4mm] &\,\hspace{2pt}+\bigg(\frac1{2\cdot4^2}-\frac1{3\cdot4^3}\bigg)\\[4mm] &\,\hspace{2pt}+\bigg(\frac1{3\cdot4^3}-\frac1{4\cdot4^4}\bigg)+\cdots\\[4mm] &\,\hspace{4pt}+\bigg(\frac1{(N-2)4^{N-2}}-\frac1{(N-1)4^{N-1}}\bigg)\\[4mm] &\,\hspace{6pt}+\bigg(\frac1{(N-1)4^{N-1}}-\frac1{N4^N}\bigg)\\[4mm] &\,\hspace{8pt}+\bigg(\frac1{N4^N}-\frac1{(N+1)4^{N+1}}\bigg)\\[4mm] =&\,\boxed{\frac14-\frac1{4^{N+1}(N+1)}}. \end{align*}

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

(c)

解法一

思路

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承接有限和，余项分母随N趋于无穷，因而其倒数趋于零。

答题过程

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By (b),

lim⁡N→∞SN=14−lim⁡N→∞14N+1(N+1)=14.\lim_{N\to\infty}S_N =\frac14-\lim_{N\to\infty}\frac1{4^{N+1}(N+1)} =\boxed{\frac14}.