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IAL 2024 Jan Q3

A Level / Edexcel / FP2

IAL 2024 Jan Paper · Question 3

题目

Problem

(a) Show that for r⩾1r \geqslant 1

rr(r+1)+r(r−1)=A(r(r+1)−r(r−1))\begin{align*} \frac{r}{\sqrt{r(r + 1)} + \sqrt{r(r - 1)}} ={}& A\left(\sqrt{r(r + 1)} - \sqrt{r(r - 1)}\right) \end{align*}

where AA is a constant to be determined.

(2)

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

∑r=1nrr(r+1)+r(r−1)\begin{align*} \sum_{r=1}^{n} \frac{r}{\sqrt{r(r + 1)} + \sqrt{r(r - 1)}} \end{align*}
(3)

(c) Determine, as a surd in simplest form, the constant kk such that

∑r=1nkrr(r+1)+r(r−1)=∑r=1nr\begin{align*} \sum_{r=1}^{n} \frac{kr}{\sqrt{r(r + 1)} + \sqrt{r(r - 1)}} ={}& \sqrt{\sum_{r=1}^{n} r} \end{align*}
(2)

解答

(a)

解法一

思路

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目标右边是两个根式相减，所以对左边分母乘以共轭式。分母平方差会变成 r(r+1)−r(r−1)=2rr(r+1)-r(r-1)=2r，于是常数 AA 就出来了。

答题过程

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Rationalise the denominator:

rr(r+1)+r(r−1)=rr(r+1)+r(r−1)⋅r(r+1)−r(r−1)r(r+1)−r(r−1)=r(r(r+1)−r(r−1))r(r+1)−r(r−1)=r(r(r+1)−r(r−1))2r=12(r(r+1)−r(r−1))\begin{align*} \frac{r}{\sqrt{r(r+1)}+\sqrt{r(r-1)}} ={}& \frac{r}{\sqrt{r(r+1)}+\sqrt{r(r-1)}} \cdot \frac{\sqrt{r(r+1)}-\sqrt{r(r-1)}} {\sqrt{r(r+1)}-\sqrt{r(r-1)}}\\[4mm] ={}& \frac{r\left(\sqrt{r(r+1)}-\sqrt{r(r-1)}\right)} {r(r+1)-r(r-1)}\\[4mm] ={}& \frac{r\left(\sqrt{r(r+1)}-\sqrt{r(r-1)}\right)} {2r}\\[4mm] ={}& \frac12\left(\sqrt{r(r+1)}-\sqrt{r(r-1)}\right) \end{align*}

Therefore

A=12\begin{align*} \boxed{A=\frac12} \end{align*}

(b)

解法一

思路

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使用 (a) 后，和式变成相邻根式相减。写出前几项和最后几项，会看到从 2\sqrt{2}、6\sqrt6 等中间项全部抵消，只留下最后一项。

答题过程

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Using part (a),

∑r=1nrr(r+1)+r(r−1)=12∑r=1n(r(r+1)−r(r−1))\begin{align*} \sum_{r=1}^{n} \frac{r}{\sqrt{r(r+1)}+\sqrt{r(r-1)}} ={}& \frac12\sum_{r=1}^{n} \left(\sqrt{r(r+1)}-\sqrt{r(r-1)}\right) \end{align*}

Write out the terms:

12∑r=1n(r(r+1)−r(r−1))=12(2−0) +12(6−2) +12(12−6) +⋯ +12(n(n+1)−n(n−1))\begin{align*} \frac12\sum_{r=1}^{n} \left(\sqrt{r(r+1)}-\sqrt{r(r-1)}\right) ={}& \frac12(\sqrt2-0)\\[4mm] &\,\hspace{2pt}+\frac12(\sqrt6-\sqrt2)\\[4mm] &\,\hspace{4pt}+\frac12(\sqrt{12}-\sqrt6)\\[4mm] &\,\hspace{6pt}+\cdots\\[4mm] &\,\hspace{8pt}+\frac12\left(\sqrt{n(n+1)}-\sqrt{n(n-1)}\right) \end{align*}

All intermediate terms cancel, so

∑r=1nrr(r+1)+r(r−1)=12n(n+1)\begin{align*} \sum_{r=1}^{n} \frac{r}{\sqrt{r(r+1)}+\sqrt{r(r-1)}} ={}& \frac12\sqrt{n(n+1)} \end{align*}

(c)

解法一

思路

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左边就是 kk 倍 (b) 的结果。右边用 ∑r=1nr=12n(n+1)\sum_{r=1}^{n}r=\frac12n(n+1)，然后比较 n(n+1)\sqrt{n(n+1)} 的系数。

答题过程

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From part (b),

∑r=1nkrr(r+1)+r(r−1)=k2n(n+1)\begin{align*} \sum_{r=1}^{n} \frac{kr}{\sqrt{r(r+1)}+\sqrt{r(r-1)}} ={}& \frac{k}{2}\sqrt{n(n+1)} \end{align*}

Also,

∑r=1nr=12n(n+1)=12n(n+1)\begin{align*} \sqrt{\sum_{r=1}^{n}r} ={}& \sqrt{\frac12n(n+1)}\\[4mm] ={}& \frac{1}{\sqrt2}\sqrt{n(n+1)} \end{align*}

Therefore

k2=12k=22k=2\begin{align*} \frac{k}{2}={}&\frac{1}{\sqrt2}\\[4mm] k={}&\frac{2}{\sqrt2}\\[4mm] k={}&\boxed{\sqrt2} \end{align*}