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FP2 2023 June Q1

A Level / Edexcel / FP2

IAL 2023 June Paper · Question 1

题目

Problem

1.

In this question you must show all stages of your working.

Solutions relying on calculator technology are not acceptable.

(a) Show that, for r⩾2r \geqslant 2

2r+r−2= r−r−2\begin{align*} \frac{2}{\sqrt{r} + \sqrt{r - 2}} =&\, \sqrt{r} - \sqrt{r - 2} \end{align*}
(2)

(b) Hence use the method of differences to determine

∑r=2n2r+r−2\begin{align*} \sum_{r=2}^{n} \frac{2}{\sqrt{r} + \sqrt{r - 2}} \end{align*}

giving your answer in simplest form.

(3)

(c) Hence show that

∑r=4502r+r−2= A+B2+C3\begin{align*} \sum_{r=4}^{50} \frac{2}{\sqrt{r} + \sqrt{r - 2}} =&\, A + B\sqrt{2} + C\sqrt{3} \end{align*}

where AA , BB and CC are integers to be determined.

(2)
题目中文翻译

本题必须展示所有运算步骤。

依赖计算器技术的解答不予接受。

(a) 证明：当 r⩾2r\geqslant2 时，

2r+r−2= r−r−2. \begin{align*} \frac2{\sqrt r+\sqrt{r-2}}=&\,\sqrt r-\sqrt{r-2}. \end{align*}
(2)

(b) 由此利用逐项相消法求

∑r=2n2r+r−2, \begin{align*} \sum_{r=2}^n\frac2{\sqrt r+\sqrt{r-2}}, \end{align*}

并将答案化为最简形式。

(3)

(c) 由此证明

∑r=4502r+r−2= A+B2+C3, \begin{align*} \sum_{r=4}^{50}\frac2{\sqrt r+\sqrt{r-2}} =&\,A+B\sqrt2+C\sqrt3, \end{align*}

其中 AA、BB、CC 为待求的整数。

(2)

解答

(a)

解法一

思路

展开

分母是两个根式相加，目标式是两个根式相减，所以自然想到乘以共轭式。这里 r⩾2r\geqslant 2，因此 r\sqrt{r} 和 r−2\sqrt{r-2} 都有意义。

答题过程

展开

Rationalise the denominator:

 2r+r−2= 2r+r−2⋅r−r−2r−r−2= 2(r−r−2)(r)2−(r−2)2= 2(r−r−2)r−(r−2)= 2(r−r−2)2= r−r−2\begin{align*} &\,\frac{2}{\sqrt{r}+\sqrt{r-2}}\\[2mm] =&\, \frac{2}{\sqrt{r}+\sqrt{r-2}} \cdot \frac{\sqrt{r}-\sqrt{r-2}}{\sqrt{r}-\sqrt{r-2}}\\[4mm] =&\, \frac{2(\sqrt{r}-\sqrt{r-2})} {(\sqrt{r})^2-(\sqrt{r-2})^2}\\[4mm] =&\, \frac{2(\sqrt{r}-\sqrt{r-2})}{r-(r-2)}\\[4mm] =&\, \frac{2(\sqrt{r}-\sqrt{r-2})}{2}\\[4mm] =&\, \sqrt{r}-\sqrt{r-2} \end{align*}

This proves the required identity.

(b)

解法一

思路

展开

利用 (a) 把每一项改写成两个根式相减。展开前几项和最后几项后，会看到中间项成对抵消，这就是 method of differences。

答题过程

展开

Using the result from part (a),

∑r=2n2r+r−2= ∑r=2n(r−r−2)\begin{align*} \sum_{r=2}^{n} \frac{2}{\sqrt{r}+\sqrt{r-2}} =&\, \sum_{r=2}^{n}(\sqrt{r}-\sqrt{r-2}) \end{align*}

Write out the important terms:

∑r=2n(r−r−2)= (2−0) +(3−1) +(4−2) +(5−3) +⋯ +(n−1−n−3) +(n−n−2)\begin{align*} \sum_{r=2}^{n}(\sqrt{r}-\sqrt{r-2}) =&\, (\sqrt{2}-\sqrt{0})\\[4mm] &\,\hspace{2pt}+(\sqrt{3}-\sqrt{1})\\[4mm] &\,\hspace{4pt}+(\sqrt{4}-\sqrt{2})\\[4mm] &\,\hspace{6pt}+(\sqrt{5}-\sqrt{3})\\[4mm] &\,\hspace{8pt}+\cdots\\[4mm] &\,\hspace{10pt}+(\sqrt{n-1}-\sqrt{n-3})\\[4mm] &\,\hspace{12pt}+(\sqrt{n}-\sqrt{n-2}) \end{align*}

The middle terms cancel in pairs. The surviving terms are

∑r=2n2r+r−2= n+n−1−1−0= n+n−1−1\begin{align*} \sum_{r=2}^{n} \frac{2}{\sqrt{r}+\sqrt{r-2}} =&\, \sqrt{n}+\sqrt{n-1}-\sqrt{1}-\sqrt{0}\\[4mm] =&\, \sqrt{n}+\sqrt{n-1}-1 \end{align*}

(c)

解法一

思路

展开

从 r=4r=4 到 5050 的和，可以用 (b) 中从 r=2r=2 开始的结果相减得到。要减掉 r=2r=2 和 r=3r=3 两项，也就是用 S50−S3S_{50}-S_3。

答题过程

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Let

Sn= ∑r=2n2r+r−2=n+n−1−1\begin{align*} S_n =&\, \sum_{r=2}^{n} \frac{2}{\sqrt{r}+\sqrt{r-2}} =\sqrt{n}+\sqrt{n-1}-1 \end{align*}

Therefore,

∑r=4502r+r−2= S50−S3= (50+49−1)−(3+2−1)= (52+7−1)−(3+2−1)= 52+6−3−2+1= 7+42−3\begin{align*} \sum_{r=4}^{50} \frac{2}{\sqrt{r}+\sqrt{r-2}} =&\, S_{50}-S_3\\[4mm] =&\, (\sqrt{50}+\sqrt{49}-1) -(\sqrt{3}+\sqrt{2}-1)\\[4mm] =&\, (5\sqrt{2}+7-1)-(\sqrt{3}+\sqrt{2}-1)\\[4mm] =&\, 5\sqrt{2}+6-\sqrt{3}-\sqrt{2}+1\\[4mm] =&\, 7+4\sqrt{2}-\sqrt{3} \end{align*}

Hence

A=7,B=4,C=−1\begin{align*} A=7,\qquad B=4,\qquad C=-1 \end{align*}