IAL 2023 June Q1 | 國際數學站

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

Solutions relying on calculator technology are not acceptable.

(a) Show that, for $r \geqslant 2$

\begin{align*} \frac{2}{\sqrt{r} + \sqrt{r - 2}} =\,& \sqrt{r} - \sqrt{r - 2}\\[2mm] \end{align*}

(2)

(b) Hence use the method of differences to determine

\begin{align*} \sum_{r=2}^{n} \frac{2}{\sqrt{r} + \sqrt{r - 2}}\\[2mm] \end{align*}

giving your answer in simplest form.

(3)

\begin{align*} \sum_{r=4}^{50} \frac{2}{\sqrt{r} + \sqrt{r - 2}} =\,& A + B\sqrt{2} + C\sqrt{3}\\[2mm] \end{align*}

where $A$ , $B$ and $C$ are integers to be determined.

(2)

解答

(a)

To show the given identity, we rationalise the denominator by multiplying the numerator and denominator by the conjugate $\sqrt{r} - \sqrt{r-2}$ :

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

(b)

Using the result from (a), we can rewrite the sum:

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

Writing out the terms to show the method of differences:

\begin{align*} r=2: \quad &\,\sqrt{2} - \sqrt{0}\\[4mm] r=3: \quad &\,\sqrt{3} - \sqrt{1}\\[4mm] r=4: \quad &\,\sqrt{4} - \sqrt{2}\\[4mm] &\,\vdots\\[4mm] r=n-1: \quad &\,\sqrt{n-1} - \sqrt{n-3}\\[4mm] r=n: \quad &\,\sqrt{n} - \sqrt{n-2} \end{align*}

Adding these terms together, all intermediate terms cancel diagonally, leaving the first two negative terms and the last two positive terms:

\begin{align*} \text{Sum} = &\,\sqrt{n} + \sqrt{n-1} - \sqrt{1} - \sqrt{0}\\[4mm] = &\,\sqrt{n} + \sqrt{n-1} - 1 \end{align*}

(c)

We need to evaluate the sum from $r=4$ to $50$ . We can express this using the result from (b):

\begin{align*} \sum_{r=4}^{50} \frac{2}{\sqrt{r}+\sqrt{r-2}} = &\,\sum_{r=2}^{50} \frac{2}{\sqrt{r}+\sqrt{r-2}} - \sum_{r=2}^{3} \frac{2}{\sqrt{r}+\sqrt{r-2}} \end{align*}

Let $f(k) = \sqrt{k} + \sqrt{k-1} - 1$ . For $n=50$ :

\begin{align*} f(50) = &\,\sqrt{50} + \sqrt{49} - 1\\[4mm] = &\,5\sqrt{2} + 7 - 1\\[4mm] = &\,6 + 5\sqrt{2} \end{align*}

For $n=3$ :

\begin{align*} f(3) = &\,\sqrt{3} + \sqrt{2} - 1 \end{align*}

Subtracting $f(3)$ from $f(50)$ :

\begin{align*} \text{Sum} = &\,(6 + 5\sqrt{2}) - (\sqrt{3} + \sqrt{2} - 1)\\[4mm] = &\,6 + 5\sqrt{2} - \sqrt{3} - \sqrt{2} + 1\\[4mm] = &\,7 + 4\sqrt{2} - \sqrt{3} \end{align*}

Comparing this to $A + B\sqrt{2} + C\sqrt{3}$ , we find that $A = 7$ , $B = 4$ , and $C = -1$ .