IAL 2025 Jan Q4 | 國際數學站

(a) Express $\frac{r + 4}{r(r + 1)(r + 2)}$ in partial fractions.

(4)

(b) Hence, using the method of differences, show that

\begin{align*} \sum_{r=1}^{n} \frac{r + 4}{r(r + 1)(r + 2)} = \frac{n(Pn + Q)}{2(n + R)(n + S)} \end{align*}

where $P$ , $Q$ , $R$ and $S$ are integers to be found.

(5)

解答

(a)

Let

\begin{align*} \frac{r+4}{r(r+1)(r+2)} = &\,\frac{A}{r}+\frac{B}{r+1}+\frac{C}{r+2} \end{align*}

Then

\begin{align*} r+4 = &\,A(r+1)(r+2)+Br(r+2)+Cr(r+1) \end{align*}

Putting $r=0,-1,-2$ gives

\begin{align*} A=2,\qquad B=-3,\qquad C=1 \end{align*}

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

(b)

Hence

\begin{align*} \sum_{r=1}^{n}\frac{r+4}{r(r+1)(r+2)} = &\,\sum_{r=1}^{n}\left(\frac{2}{r(r+1)}-\frac{1}{(r+1)(r+2)}\right)\\[4mm] = &\,2\sum_{r=1}^{n}\left(\frac{1}{r}-\frac{1}{r+1}\right) - \sum_{r=1}^{n}\left(\frac{1}{r+1}-\frac{1}{r+2}\right)\\[4mm] = &\,2\left(1-\frac{1}{n+1}\right) -\left(\frac12-\frac{1}{n+2}\right)\\[4mm] = &\,\frac{n(3n+7)}{2(n+1)(n+2)} \end{align*}

Therefore

\begin{align*} P=3,\qquad Q=7,\qquad R=1,\qquad S=2 \end{align*}