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IAL 2025 June A Q5

A Level / Edexcel / FP2

IAL 2025 June A Paper · Question 5

题目

Problem

In this question you must show all stages of your working.
Solutions relying entirely on calculator technology are not acceptable.

(a) Express

1+22r+5\begin{align*} 1+\frac{2}{2r+5} \end{align*}

as a single fraction in simplest form.

(1)

(b) Hence use the method of differences to determine an expression for

∑r=1nlog⁡3(1+22r+5)\begin{align*} \sum_{r=1}^{n}\log_3\left(1+\frac{2}{2r+5}\right) \end{align*}

giving the answer in the form log⁡3(f(n))\log_3(f(n)) where ff is a function to be found.

(3)

(c) Hence determine the value of nn for which

∑r=n+210nlog⁡3(1+22r+5)=2\begin{align*} \sum_{r=n+2}^{10n}\log_3\left(1+\frac{2}{2r+5}\right)=2 \end{align*}
(4)

解答

(a)

解法一

思路

展开

把 11 写成同分母的分式即可。这个结果会让后面的对数变成可以相消的形式。

#### 答题过程
展开 1+22r+5= 2r+52r+5+22r+5= 2r+72r+5\begin{align*} 1+\frac{2}{2r+5} =&\,\frac{2r+5}{2r+5}+\frac{2}{2r+5}\\[4mm] =&\,\frac{2r+7}{2r+5} \end{align*}

(b)

解法一

思路

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先用 log⁡(a/b)=log⁡a−log⁡b\log(a/b)=\log a-\log b，然后把前几项和最后一项写出来。中间的 log⁡39,log⁡311,…\log_3 9,\log_3 11,\ldots 会一正一负抵消。

答题过程

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Using part (a),

∑r=1nlog⁡3(1+22r+5)= ∑r=1nlog⁡3(2r+72r+5)= ∑r=1n(log⁡3(2r+7)−log⁡3(2r+5))\begin{align*} \sum_{r=1}^{n}\log_3\left(1+\frac{2}{2r+5}\right) =&\,\sum_{r=1}^{n}\log_3\left(\frac{2r+7}{2r+5}\right)\\[4mm] =&\,\sum_{r=1}^{n} \left(\log_3(2r+7)-\log_3(2r+5)\right) \end{align*}

Writing out the terms,

r=1:log⁡39−log⁡37r=2:log⁡311−log⁡39r=3:log⁡313−log⁡311⋮r=n:log⁡3(2n+7)−log⁡3(2n+5)\begin{align*} r=1:\quad &\log_3 9-\log_3 7\\[2mm] r=2:\quad &\log_3 11-\log_3 9\\[2mm] r=3:\quad &\log_3 13-\log_3 11\\[2mm] &\vdots\\[2mm] r=n:\quad &\log_3(2n+7)-\log_3(2n+5) \end{align*}

After cancellation,

∑r=1nlog⁡3(1+22r+5)= log⁡3(2n+7)−log⁡37= log⁡3(2n+77)\begin{align*} \sum_{r=1}^{n}\log_3\left(1+\frac{2}{2r+5}\right) =&\,\log_3(2n+7)-\log_3 7\\[4mm] =&\,\log_3\left(\frac{2n+7}{7}\right) \end{align*}

解法二

思路

展开

也可以先把对数和合并成一个乘积。分子分母会连锁约掉，这其实是同一个望远镜结构，只是写成乘法形式。

答题过程

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Using part (a),

∑r=1nlog⁡3(1+22r+5)= log⁡3(97)+log⁡3(119) +log⁡3(1311)+⋯+log⁡3(2n+72n+5)= log⁡3(97⋅119⋅1311⋯2n+72n+5)= log⁡3(2n+77)\begin{align*} \sum_{r=1}^{n}\log_3\left(1+\frac{2}{2r+5}\right) =&\,\log_3\left(\frac{9}{7}\right) +\log_3\left(\frac{11}{9}\right)\\[2mm] &\,\hspace{2pt}+\log_3\left(\frac{13}{11}\right) +\cdots +\log_3\left(\frac{2n+7}{2n+5}\right)\\[4mm] =&\,\log_3\left( \frac{9}{7}\cdot\frac{11}{9}\cdot \frac{13}{11}\cdots \frac{2n+7}{2n+5} \right)\\[4mm] =&\,\log_3\left(\frac{2n+7}{7}\right) \end{align*}

(c)

解法一

思路

展开

利用 (b) 的结果，把从 r=n+2r=n+2 到 10n10n 的和写成「前 10n10n 项」减去「前 n+1n+1 项」。注意下限是 n+2n+2，所以要减到 n+1n+1。

答题过程

展开

From part (b), define

Sk= ∑r=1klog⁡3(1+22r+5)= log⁡3(2k+77)\begin{align*} S_k =&\,\sum_{r=1}^{k}\log_3\left(1+\frac{2}{2r+5}\right)\\[4mm] =&\,\log_3\left(\frac{2k+7}{7}\right) \end{align*}

Then

∑r=n+210nlog⁡3(1+22r+5)= S10n−Sn+1= log⁡3(20n+77)−log⁡3(2n+97)= log⁡3(20n+72n+9)\begin{align*} \sum_{r=n+2}^{10n}\log_3\left(1+\frac{2}{2r+5}\right) =&\,S_{10n}-S_{n+1}\\[4mm] =&\,\log_3\left(\frac{20n+7}{7}\right) -\log_3\left(\frac{2n+9}{7}\right)\\[4mm] =&\,\log_3\left(\frac{20n+7}{2n+9}\right) \end{align*}

So

log⁡3(20n+72n+9)= 220n+72n+9= 3220n+7= 9(2n+9)20n+7= 18n+812n= 74n= 37\begin{align*} \log_3\left(\frac{20n+7}{2n+9}\right) =&\,2\\[4mm] \frac{20n+7}{2n+9} =&\,3^2\\[4mm] 20n+7 =&\,9(2n+9)\\[4mm] 20n+7 =&\,18n+81\\[4mm] 2n =&\,74\\[4mm] n =&\,37 \end{align*}