Use standard results from the list of formulae (MF19) to show that
$$\sum _ { r = 1 } ^ { N } r ( r + 1 ) ( 3 r + 4 ) = \frac { 1 } { 12 } N ( N + 1 ) ( N + 2 ) ( 9 N + 19 )$$
Express \(\frac { 3 r + 4 } { r ( r + 1 ) }\) in partial fractions and hence use the method of differences to find
$$\sum _ { r = 1 } ^ { N } \frac { 3 r + 4 } { r ( r + 1 ) } \left( \frac { 1 } { 4 } \right) ^ { r + 1 }$$
in terms of \(N\).
Deduce the value of \(\sum _ { r = 1 } ^ { \infty } \frac { 3 r + 4 } { r ( r + 1 ) } \left( \frac { 1 } { 4 } \right) ^ { r + 1 }\).