Standard +0.8 This is a structured method of differences question requiring algebraic manipulation to verify the given identity, telescoping series summation, and ratio calculation. While the identity is provided (reducing discovery difficulty), students must handle partial fractions in reverse, recognize the telescoping pattern, and apply the formula correctly. This is moderately challenging for Further Maths, requiring careful algebraic work across multiple parts but following a standard technique once the pattern is identified.
9. (a) Given that \(A _ { r } = \frac { 1 } { r + 1 } - \frac { 2 } { r + 2 } + \frac { 1 } { r + 3 }\), show that \(A _ { r }\) can be expressed as \(\frac { 2 } { ( r + 1 ) ( r + 2 ) ( r + 3 ) }\).
(b) Hence, show that \(\sum _ { r = 1 } ^ { n } \frac { 2 } { ( r + 1 ) ( r + 2 ) ( r + 3 ) } = \frac { 1 } { 6 } - \frac { 1 } { ( n + 2 ) ( n + 3 ) }\).
(c) Find the ratio of \(\sum _ { r = 1 } ^ { 5 } A _ { r } : \sum _ { r = 1 } ^ { 10 } A _ { r }\), giving your answer in its simplest form.
9. (a) Given that $A _ { r } = \frac { 1 } { r + 1 } - \frac { 2 } { r + 2 } + \frac { 1 } { r + 3 }$, show that $A _ { r }$ can be expressed as $\frac { 2 } { ( r + 1 ) ( r + 2 ) ( r + 3 ) }$.\\
(b) Hence, show that $\sum _ { r = 1 } ^ { n } \frac { 2 } { ( r + 1 ) ( r + 2 ) ( r + 3 ) } = \frac { 1 } { 6 } - \frac { 1 } { ( n + 2 ) ( n + 3 ) }$.\\
(c) Find the ratio of $\sum _ { r = 1 } ^ { 5 } A _ { r } : \sum _ { r = 1 } ^ { 10 } A _ { r }$, giving your answer in its simplest form.
\hfill \mbox{\textit{WJEC Further Unit 1 2022 Q9 [9]}}