I3.2 Determine the value of $\beta$ if $\frac{\beta}{1 \times 2 \times 3}+\frac{\beta}{2 \times 3 \times 4}+\cdots+\frac{\beta}{8 \times 9 \times 10}=\alpha$.
$$
\frac{1}{r(r+1)}-\frac{1}{(r+1)(r+2)}=\frac{(r+2)-r}{r(r+1)(r+2)}=2 \cdot \frac{1}{r(r+1)(r+2)}
$$

Put $r=1, \frac{1}{1 \times 2}-\frac{1}{2 \times 3}=2 \cdot \frac{1}{1 \times 2 \times 3}$
Put $r=2, \frac{1}{2 \times 3}-\frac{1}{3 \times 4}=2 \cdot \frac{1}{2 \times 3 \times 4}$
Put $r=8, \frac{1}{8 \times 9}-\frac{1}{9 \times 10}=2 \cdot \frac{1}{8 \times 9 \times 10}$
Add these equations together and multiply both sides by $\beta$ and divide by 2 :
$$
\begin{array}{l}
\frac{\beta}{2}\left[\frac{1}{2}-\frac{1}{9 \times 10}\right]=\frac{\beta}{1 \times 2 \times 3}+\frac{\beta}{2 \times 3 \times 4}+\cdots+\frac{\beta}{8 \times 9 \times 10}=\alpha=11 \\
\beta=45
\end{array}
$$