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FP2 2024 Jan Q3

A Level / Edexcel / FP2

IAL 2024 Jan Paper · Question 3

题目

Problem

3.

(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\bigg(\sqrt{r(r + 1)} - \sqrt{r(r - 1)}\bigg) \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)
题目中文翻译

(a) 证明：当 r⩾1r\geqslant1 时，

 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)}}\\[2mm] =&\,A\big(\sqrt{r(r+1)}-\sqrt{r(r-1)}\big), \end{align*}

其中 AA 为待求的常数。

(2)

(b) 由此利用逐项相消法，求下列和式的化简表达式：

∑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) 求使下式成立的常数 kk，将答案写成最简根式：

∑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)}}\\[2mm] =&\, \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\bigg(\sqrt{r(r+1)}-\sqrt{r(r-1)}\bigg)} {r(r+1)-r(r-1)}\\[4mm] =&\, \frac{r\bigg(\sqrt{r(r+1)}-\sqrt{r(r-1)}\bigg)} {2r}\\[4mm] =&\, \frac12\bigg(\sqrt{r(r+1)}-\sqrt{r(r-1)}\bigg) \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)}}\\[2mm] =&\, \frac12\sum_{r=1}^{n} \bigg(\sqrt{r(r+1)}-\sqrt{r(r-1)}\bigg) \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−1)(n−2)) +12(n(n+1)−n(n−1))\begin{align*} &\,\frac12\sum_{r=1}^{n} \bigg(\sqrt{r(r+1)}-\sqrt{r(r-1)}\bigg)\\[2mm] =&\, \frac12(\sqrt2-0)\\[4mm] &\,\hspace{2pt}+\frac12(\sqrt6-\sqrt2)\\[4mm] &\,\hspace{4pt}+\frac12(\sqrt{12}-\sqrt6)\\[4mm] &\,\hspace{6pt}+\cdots\\[4mm] &\,\hspace{8pt}+\frac12\bigg(\sqrt{n(n-1)}-\sqrt{(n-1)(n-2)}\bigg)\\[4mm] &\,\hspace{10pt}+\frac12\bigg(\sqrt{n(n+1)}-\sqrt{n(n-1)}\bigg) \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*}