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FP2 2025 June A Q5

A Level / Edexcel / FP2

IAL 2025 June A Paper · Question 5

题目

Problem

5.

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)
题目中文翻译

本题必须展示所有解题步骤。不接受完全依赖计算器技术的解答。

(a) 将 1+22r+51+\dfrac{2}{2r+5} 化为最简的单个分式。

(1)

(b) 由此利用差分相消法求

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

的表达式，并将答案写成 log⁡3(f(n))\log_3(f(n)) 的形式，其中 ff 是需要求出的函数。

(3)

(c) 由此求满足

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

的 nn 值。

(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

展开

先用 log⁡(a/b)=log⁡a−log⁡b\log(a/b)=\log a-\log b，然后把前三项和最后两项写出来。中间的 log⁡39,log⁡311,…,log⁡3(2n+5)\log_3 9,\log_3 11,\ldots,\log_3(2n+5) 会一正一负抵消，只留下最后一个正项与第一个负项。

答题过程

展开

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\Big(1+\frac{2}{2r+5}\Big) =&\,\sum_{r=1}^{n}\log_3\Big(\frac{2r+7}{2r+5}\Big)\\[4mm] =&\,\sum_{r=1}^{n} \Big(\log_3(2r+7)-\log_3(2r+5)\Big) \end{align*}

Writing out the terms,

r=1:log⁡39−log⁡37r=2:log⁡311−log⁡39r=3:log⁡313−log⁡311⋮r=n−1:log⁡3(2n+5)−log⁡3(2n+3)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-1:\quad &\log_3(2n+5)-\log_3(2n+3)\\[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\Big(1+\frac{2}{2r+5}\Big) =&\,\log_3(2n+7)-\log_3 7\\[4mm] =&\,\log_3\Big(\frac{2n+7}{7}\Big) \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

展开

Using part (a),

∑r=1nlog⁡3(1+22r+5)= log⁡3(97)+log⁡3(119) +log⁡3(1311)+⋯+log⁡3(2n+52n+3) +log⁡3(2n+72n+5)= log⁡3(97⋅119⋅1311⋯ ⋅2n+52n+3⋅2n+72n+5)= log⁡3(2n+77)\begin{align*} \sum_{r=1}^{n}\log_3\Big(1+\frac{2}{2r+5}\Big) =&\,\log_3\Big(\frac{9}{7}\Big) +\log_3\Big(\frac{11}{9}\Big)\\[2mm] &\,\hspace{2pt}+\log_3\Big(\frac{13}{11}\Big) +\cdots +\log_3\Big(\frac{2n+5}{2n+3}\Big)\\[2mm] &\,\hspace{4pt} +\log_3\Big(\frac{2n+7}{2n+5}\Big)\\[4mm] =&\,\log_3\Big( \frac{9}{7}\cdot\frac{11}{9}\cdot \frac{13}{11}\cdots\\[2mm] &\,\hspace{2pt}\cdot\frac{2n+5}{2n+3}\cdot \frac{2n+7}{2n+5} \Big)\\[4mm] =&\,\log_3\Big(\frac{2n+7}{7}\Big) \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\Big(1+\frac{2}{2r+5}\Big)\\[4mm] =&\,\log_3\Big(\frac{2k+7}{7}\Big) \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\Big(1+\frac{2}{2r+5}\Big) =&\,S_{10n}-S_{n+1}\\[4mm] =&\,\log_3\Big(\frac{20n+7}{7}\Big) -\log_3\Big(\frac{2n+9}{7}\Big)\\[4mm] =&\,\log_3\Big(\frac{20n+7}{2n+9}\Big) \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\Big(\frac{20n+7}{2n+9}\Big) =&\,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*}