IAL 2020 Oct Q2 | 國際數學站

(a) Write

\begin{align*} \frac{3r + 1}{r(r - 1)(r + 1)} \end{align*}

in partial fractions.

(2)

(b) Hence find

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

giving your answer in the form

\begin{align*} \frac{an^2 + bn + c}{2n(n + 1)} \end{align*}

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

(5)

\begin{align*} \sum_{r=15}^{20} \frac{3r + 1}{r(r - 1)(r + 1)} \end{align*}

(2)

解答

(a)

\begin{align*} \frac{3r + 1}{r(r - 1)(r + 1)} = &\,\frac{A}{r} + \frac{B}{r - 1} + \frac{C}{r + 1}\\[4mm] 3r + 1 = &\,A(r - 1)(r + 1) + Br(r + 1)\\[4mm] &\,\hspace{2pt}+ Cr(r - 1)\\[4mm] = &\,(A + B + C)r^2 + (B - C)r - A \end{align*}

\begin{align*} A + B + C =&\, 0\\[4mm] B - C =&\, 3\\[4mm] -A =&\, 1 \end{align*}

Hence $A = -1$ , $B = 2$ and $C = -1$ .

(b)

\begin{align*} S_n = &\,\sum_{r=2}^{n} \frac{3r + 1}{r(r - 1)(r + 1)}\\[4mm] = &\,\sum_{r=2}^{n} \left(\frac{2}{r - 1} - \frac{1}{r} - \frac{1}{r + 1}\right)\\[4mm] = &\,\left(2 - \frac{1}{2} - \frac{1}{3}\right) + \left(1 - \frac{1}{3} - \frac{1}{4}\right)\\[4mm] &\,\hspace{2pt}+ \cdots + \left(\frac{2}{n - 1} - \frac{1}{n} - \frac{1}{n + 1}\right)\\[4mm] &\,\hspace{2pt}\colorbox{aqua}{middle terms cancel}\\[4mm] = &\,2\left(1 - \frac{1}{n}\right) - \left(\frac{1}{n + 1} - \frac{1}{2}\right)\\[4mm] = &\,\frac{5n^2 - n - 4}{2n(n + 1)} \end{align*}

(c)

Let

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

then

\begin{align*} &\,\sum_{r=15}^{20} \frac{3r + 1}{r(r - 1)(r + 1)}\\[4mm] = &\,S_{20} - S_{14}\\[4mm] = &\,\frac{5 \cdot 20^2 - 20 - 4}{2 \cdot 20 \cdot 21} - \frac{5 \cdot 14^2 - 14 - 4}{2 \cdot 14 \cdot 15}\\[4mm] = &\,\frac{13}{210} \end{align*}