By considering \(( r + 1 ) ^ { 2 } - r ^ { 2 }\), use the method of differences to prove that
$$\sum _ { r = 1 } ^ { n } r = \frac { 1 } { 2 } n ( n + 1 )$$
Given that \(\sum _ { \mathrm { r } = 1 } ^ { \mathrm { n } } ( \mathrm { r } + \mathrm { a } ) = \mathrm { n }\), find \(a\) in terms of \(n\).