- (a) Show that
$$\frac { 1 } { ( r + 1 ) ( r + 2 ) ( r + 3 ) } \equiv \frac { 1 } { 2 ( r + 1 ) ( r + 2 ) } - \frac { 1 } { 2 ( r + 2 ) ( r + 3 ) }$$
(b) Hence, or otherwise, find
$$\sum _ { r = 1 } ^ { n } \frac { 1 } { ( r + 1 ) ( r + 2 ) ( r + 3 ) }$$
giving your answer as a single fraction in its simplest form.