IAL 2024 June Q3 | 國際數學站

(a) Express

\begin{align*} \frac{1}{(n + 3)(n + 5)} \end{align*}

in partial fractions.

(2)

(b) Hence, using the method of differences, show that for all positive integer values of $n$ ,

\begin{align*} \sum_{r=1}^{n} \frac{1}{(r + 3)(r + 5)} = \frac{n(pn + q)}{40(n + 4)(n + 5)} \end{align*}

where $p$ and $q$ are integers to be determined.

(4)

\begin{align*} \frac{1}{9 \times 11} + \frac{1}{10 \times 12} + \dots + \frac{1}{24 \times 26} \end{align*}

(2)

解答

(a)

Let the partial fractions be:

\begin{align*} \frac{1}{(n+3)(n+5)} \equiv &\,\frac{A}{n+3} + \frac{B}{n+5}\\[4mm] 1 \equiv &\,A(n+5) + B(n+3) \end{align*}

Substitute $n = -3$ :

\begin{align*} 1 = &\,A(-3+5)\\[4mm] 1 = &\,2A\\[4mm] A = &\,\frac{1}{2} \end{align*}

Substitute $n = -5$ :

\begin{align*} 1 = &\,B(-5+3)\\[4mm] 1 = &\,-2B\\[4mm] B = &\,-\frac{1}{2} \end{align*}

Therefore:

\begin{align*} \frac{1}{(n+3)(n+5)} = &\,\frac{1}{2(n+3)} - \frac{1}{2(n+5)} \end{align*}

(b)

Using the method of differences:

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

Writing out the first few and last few terms:

\begin{align*} r=1: \quad &\,\frac{1}{4} - \frac{1}{6}\\[4mm] r=2: \quad &\,\frac{1}{5} - \frac{1}{7}\\[4mm] r=3: \quad &\,\frac{1}{6} - \frac{1}{8}\\[4mm] &\,\vdots\\[4mm] r=n-1: \quad &\,\frac{1}{n+2} - \frac{1}{n+4}\\[4mm] r=n: \quad &\,\frac{1}{n+3} - \frac{1}{n+5} \end{align*}

Most terms cancel diagonally, leaving the first two positive terms and the last two negative terms:

\begin{align*} \text{Sum} = &\,\frac{1}{2} \left( \frac{1}{4} + \frac{1}{5} - \frac{1}{n+4} - \frac{1}{n+5} \right)\\[4mm] = &\,\frac{1}{2} \left( \frac{9}{20} - \frac{(n+5) + (n+4)}{(n+4)(n+5)} \right)\\[4mm] = &\,\frac{1}{2} \left( \frac{9(n+4)(n+5) - 20(2n+9)}{20(n+4)(n+5)} \right)\\[4mm] = &\,\frac{1}{40(n+4)(n+5)} \Big( 9(n^2 + 9n + 20) - 40n - 180 \Big)\\[4mm] = &\,\frac{1}{40(n+4)(n+5)} \Big( 9n^2 + 81n + 180 - 40n - 180 \Big)\\[4mm] = &\,\frac{9n^2 + 41n}{40(n+4)(n+5)}\\[4mm] = &\,\frac{n(9n + 41)}{40(n+4)(n+5)} \end{align*}

This is in the required form, with $p = 9$ and $q = 41$ .

(c)

The given sum is $\frac{1}{9 \times 11} + \frac{1}{10 \times 12} + \dots + \frac{1}{24 \times 26}$ . This corresponds to $\sum_{r=6}^{21} \frac{1}{(r+3)(r+5)}$ .

\begin{align*} \sum_{r=6}^{21} \frac{1}{(r+3)(r+5)} = &\,\sum_{r=1}^{21} \frac{1}{(r+3)(r+5)} - \sum_{r=1}^{5} \frac{1}{(r+3)(r+5)}\\[4mm] = &\,\frac{21(9(21) + 41)}{40(21+4)(21+5)} - \frac{5(9(5) + 41)}{40(5+4)(5+5)}\\[4mm] = &\,\frac{21(189 + 41)}{40(25)(26)} - \frac{5(45 + 41)}{40(9)(10)}\\[4mm] = &\,\frac{21(230)}{26000} - \frac{5(86)}{3600}\\[4mm] = &\,\frac{4830}{26000} - \frac{430}{3600}\\[4mm] = &\,\frac{483}{2600} - \frac{43}{360} \end{align*}

Finding a common denominator for the fractions:

\begin{align*} \frac{483}{2600} - \frac{43}{360} = &\,\frac{483 \times 9}{23400} - \frac{43 \times 65}{23400}\\[4mm] = &\,\frac{4347 - 2795}{23400}\\[4mm] = &\,\frac{1552}{23400}\\[4mm] = &\,\frac{194}{2925} \end{align*}