IAL 2025 June 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*} \frac{2}{r(r + 1)(r + 2)} \end{align*}

in partial fractions.

(2)

(b) Use the answer to part (a) and the method of differences to show that

\begin{align*} \sum_{r=1}^{n} \frac{2}{r(r + 1)(r + 2)} = \frac{n(n + a)}{2(n + b)(n + c)} \end{align*}

where $a$ , $b$ and $c$ are integers to be determined.

(5)

\begin{align*} \sum_{r=1}^{n} \frac{2}{r(r + 1)(r + 2)} > \frac{7}{15} \end{align*}

(3)

解答

(a)

Let:

\begin{align*} \frac{2}{r(r+1)(r+2)} = \frac{A}{r} + \frac{B}{r+1} + \frac{C}{r+2} \end{align*}

Multiplying by $r(r+1)(r+2)$ :

\begin{align*} 2 = A(r+1)(r+2) + Br(r+2) + Cr(r+1) \end{align*}

Let $r = 0$ :

\begin{align*} 2 = 2A \implies A = 1 \end{align*}

Let $r = -1$ :

\begin{align*} 2 = -B \implies B = -2 \end{align*}

Let $r = -2$ :

\begin{align*} 2 = 2C \implies C = 1 \end{align*}

Hence:

\begin{align*} \frac{2}{r(r+1)(r+2)} = \frac{1}{r} - \frac{2}{r+1} + \frac{1}{r+2} \end{align*}

(b)

Using the method of differences:

\begin{align*} \sum_{r=1}^{n} \frac{2}{r(r+1)(r+2)} = &\,\sum_{r=1}^{n} \left(\frac{1}{r} - \frac{2}{r+1} + \frac{1}{r+2}\right)\\[4mm] r=1: &\, \frac{1}{1} - \frac{2}{2} + \frac{1}{3}\\[4mm] r=2: &\, \frac{1}{2} - \frac{2}{3} + \frac{1}{4}\\[4mm] r=3: &\, \frac{1}{3} - \frac{2}{4} + \frac{1}{5}\\[4mm] &\,\cdots\\[4mm] r=n-1: &\, \frac{1}{n-1} - \frac{2}{n} + \frac{1}{n+1}\\[4mm] r=n: &\, \frac{1}{n} - \frac{2}{n+1} + \frac{1}{n+2} \end{align*}

Summing the columns, most terms cancel out:

\begin{align*} = &\,1 - \frac{2}{2} + \frac{1}{2} - \frac{2}{n+1} + \frac{1}{n+1} + \frac{1}{n+2}\\[4mm] = &\,\frac{1}{2} - \frac{1}{n+1} + \frac{1}{n+2}\\[4mm] = &\,\frac{(n+1)(n+2) - 2(n+2) + 2(n+1)}{2(n+1)(n+2)}\\[4mm] = &\,\frac{n^2 + 3n + 2 - 2n - 4 + 2n + 2}{2(n+1)(n+2)}\\[4mm] = &\,\frac{n^2 + 3n}{2(n+1)(n+2)}\\[4mm] = &\,\frac{n(n+3)}{2(n+1)(n+2)} \end{align*}

Here, $a = 3$ , $b = 1$ , and $c = 2$ .

(c)

We require:

\begin{align*} \frac{n(n+3)}{2(n+1)(n+2)} > \frac{7}{15} \end{align*}

Since $n \ge 1$ , the denominators are positive. Cross-multiplying gives:

\begin{align*} 15(n^2 + 3n) > &\,14(n+1)(n+2)\\[4mm] 15n^2 + 45n > &\,14(n^2 + 3n + 2)\\[4mm] 15n^2 + 45n > &\,14n^2 + 42n + 28\\[4mm] n^2 + 3n - 28 > &\,0\\[4mm] (n + 7)(n - 4) > &\,0 \end{align*}

Since $n$ must be positive, we need $n > 4$ . The smallest integer value of $n$ is $n = 5$ .