IAL 2025 June A Q5 | 國際數學站

In this question you must show all stages of your working. Solutions relying entirely on calculator technology are not acceptable.

(a) Express

\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

\begin{align*} \sum_{r=1}^{n} \log_3\!\left(1 + \frac{2}{2r + 5}\right) \end{align*}

giving the answer in the form $\log_3(f(n))$ where $f$ is a function to be found.

(3)

\begin{align*} \sum_{r=n+2}^{10n} \log_3\!\left(1 + \frac{2}{2r + 5}\right) = 2 \end{align*}

(4)

解答

(a)

\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)

Using the result from part (a) and properties of logarithms:

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

Writing out the terms to show the method of differences:

\begin{align*} r=1: \quad &\,\log_3(9) - \log_3(7)\\[4mm] r=2: \quad &\,\log_3(11) - \log_3(9)\\[4mm] r=3: \quad &\,\log_3(13) - \log_3(11)\\[4mm] &\,\vdots\\[4mm] r=n: \quad &\,\log_3(2n+7) - \log_3(2n+5) \end{align*}

Adding these terms together, all terms cancel diagonally except for the positive part of the last term and the negative part of the first term:

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

(c)

We need to determine the value of $n$ for which:

\begin{align*} \sum_{r=n+2}^{10n} \log_3\left(1 + \frac{2}{2r+5}\right) = 2 \end{align*}

We can express this sum as the difference of two sums starting from $r=1$ :

\begin{align*} \sum_{r=n+2}^{10n} \log_3\left(1 + \frac{2}{2r+5}\right) = &\,\sum_{r=1}^{10n} \log_3\left(1 + \frac{2}{2r+5}\right) - \sum_{r=1}^{n+1} \log_3\left(1 + \frac{2}{2r+5}\right) \end{align*}

Using the result $f(k) = \log_3\left(\frac{2k+7}{7}\right)$ from part (b):

\begin{align*} &\,\log_3\left(\frac{2(10n)+7}{7}\right) - \log_3\left(\frac{2(n+1)+7}{7}\right) = 2\\[4mm] &\,\log_3\left(\frac{20n+7}{7}\right) - \log_3\left(\frac{2n+9}{7}\right) = 2\\[4mm] &\,\log_3\left(\frac{\frac{20n+7}{7}}{\frac{2n+9}{7}}\right) = 2\\[4mm] &\,\log_3\left(\frac{20n+7}{2n+9}\right) = 2 \end{align*}

Converting from logarithmic to exponential form:

\begin{align*} \frac{20n+7}{2n+9} = &\,3^2\\[4mm] \frac{20n+7}{2n+9} = &\,9\\[4mm] 20n + 7 = &\,9(2n+9)\\[4mm] 20n + 7 = &\,18n + 81\\[4mm] 2n = &\,74\\[4mm] n = &\,37 \end{align*}