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CIE 9231 2024 November Paper 11 Q4

A Level / CIE / FP1

CIE 9231 2024 Nov Paper 11 Paper · Question 4

题目

Problem

(a) Use the method of differences to find ∑r=1n5k(5r+k)(5r+5+k)\displaystyle\sum_{r=1}^{n}\frac{5k}{(5r+k)(5r+5+k)} in terms of nn and kk, where kk is a positive constant.

(4)

It is given that ∑r=1∞5k(5r+k)(5r+5+k)=13\displaystyle\sum_{r=1}^{\infty}\frac{5k}{(5r+k)(5r+5+k)}=\frac13.

(b) Find the value of kk.

(2)

(c) Hence find ∑r=nn25k(5r+k)(5r+5+k)\displaystyle\sum_{r=n}^{n^2}\frac{5k}{(5r+k)(5r+5+k)} in terms of nn.

(2)
题目中文翻译

(a) 用差分法求 ∑r=1n5k(5r+k)(5r+5+k)\displaystyle\sum_{r=1}^{n}\frac{5k}{(5r+k)(5r+5+k)},以 nn 和 kk 表示,其中 kk 为正常数。

(4)

已知 ∑r=1∞5k(5r+k)(5r+5+k)=13\displaystyle\sum_{r=1}^{\infty}\frac{5k}{(5r+k)(5r+5+k)}=\frac13。

(b) 求 kk 的值。

(2)

(c) 由此求 ∑r=nn25k(5r+k)(5r+5+k)\displaystyle\sum_{r=n}^{n^2}\frac{5k}{(5r+k)(5r+5+k)},以 nn 表示。

(2)
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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分母相差5,拆成两项后有相邻项抵消。先保留一般的k,不提前代入特殊数值。

答题过程

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Since

5k(5r+k)(5r+5+k)=k(15r+k−15r+5+k),\frac{5k}{(5r+k)(5r+5+k)} =k\bigg(\frac1{5r+k}-\frac1{5r+5+k}\bigg),

the partial sum is

Sn= k[(15+k−110+k) +(110+k−115+k) +(115+k−120+k)+⋯ +(15n−10+k−15n−5+k) +(15n−5+k−15n+k) +(15n+k−15n+5+k)]= k(15+k−15n+5+k).\begin{align*} S_n=&\,k\bigg[\bigg(\frac1{5+k}-\frac1{10+k}\bigg)\\[4mm] &\,\hspace{2pt}+\bigg(\frac1{10+k}-\frac1{15+k}\bigg)\\[4mm] &\,\hspace{2pt}+\bigg(\frac1{15+k}-\frac1{20+k}\bigg)+\cdots\\[4mm] &\,\hspace{4pt}+\bigg(\frac1{5n-10+k}-\frac1{5n-5+k}\bigg)\\[4mm] &\,\hspace{6pt}+\bigg(\frac1{5n-5+k}-\frac1{5n+k}\bigg)\\[4mm] &\,\hspace{8pt}+\bigg(\frac1{5n+k}-\frac1{5n+5+k}\bigg)\bigg]\\[4mm] =&\,\boxed{k\bigg(\frac1{5+k}-\frac1{5n+5+k}\bigg)}. \end{align*}

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

(b)

解法一

思路

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令部分和中的n趋于无穷,再解所给和为三分之一的方程。

答题过程

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Taking the limit in (a),

k5+k=13.\frac{k}{5+k}=\frac13.

Therefore 3k=5+k3k=5+k, giving k=52\boxed{k=\frac52}, consistent with k>0k>0.

(c)

解法一

思路

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从n到n平方的求和,等于前n平方项之和减前n−1项之和;起始下标不要减错。

答题过程

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Using (a) and (b), with S0=0S_0=0,

∑r=nn25k(5r+k)(5r+5+k)= Sn2−Sn−1= k(15n+k−15n2+5+k)= 5/25n+5/2−5/25n2+15/2= 12n+1−12n2+3.\begin{align*} \sum_{r=n}^{n^2}\frac{5k}{(5r+k)(5r+5+k)} =&\,S_{n^2}-S_{n-1}\\[4mm] =&\,k\bigg(\frac1{5n+k}-\frac1{5n^2+5+k}\bigg)\\[4mm] =&\,\frac{5/2}{5n+5/2}-\frac{5/2}{5n^2+15/2}\\[4mm] =&\,\boxed{\frac1{2n+1}-\frac1{2n^2+3}}. \end{align*}