(ii) $\frac{n+1}{(n+2)(n+3)}$

(iii) $-\frac{5}{66}$

(i) $\frac{2-r}{r(r+1)(r+2)}=\frac{A-B+C}{r-r+1} r+2$

Method 1: Using cover-up method, $A=1, B=-3, C=2$
Method 2:
$\frac{2-r}{r(r+1)(r+2)}=\frac{A-B}{r+1}+\frac{C}{r+2}$
$2-r=A(r+1)(r+2)+\operatorname{Br}(r+2)+\operatorname{Cr}(r+1)$
Put $r=0: 2=2 A$, then $A=1$
Put $r=-1: 3=B(-1)(-1+2) \implies$ $B$ = $-3$
Put $r=-2: 4=C(-2)(-2+1) \implies$ $C=2$

Hence, $\frac{2-r}{r(r+1)(r+2)}$ = $\frac{1}{r} – \frac{3}{r+1}+\frac{2}{r+2}$

(ii) Method 1: From part (i),
$\sum_{r=0}^n \frac{1-r}{(r+1)(r+2)(r+3)} \quad \quad$ Let $\left.r=k-1\right)$
$=\sum_{k=1=0}^{k-1=n} \frac{1-(k-1)}{(k-1+1)(k-1+2)(k-1+3)}$
$=\sum_{k=1}^{n+1} \frac{2-k}{k(k+1)(k+2)}=\frac{n+1}{(n+2)(n+3)}$

(iii)