(related to Proposition: Sum of Arithmetic Progression)
Since \(a,b\in F\) are elements of a field \((F, +, \cdot)\), we can use the commutative law and replace \(k\) by \(n-k\), obtaining
\[S=\sum_{0\le k\le n} (a+bk)=\sum_{0\le n-k\le n} (a+b(n-k)).\]
Since both sides of the equation are equal, the associative law allows the addition of both representations of \(S\), obtaining
\[2\cdot S=\sum_{0\le k\le n} ((a+bk)+(a+bn-bk))=\sum_{0\le k\le n} (2a + bn).\]
By the distributive law we obtain a trivial sum. \[2\cdot S=(2a + bn)\sum_{0\le k\le n} 1=(2a + bn)(n+1).\]
Dividing by \(2\) we obtain
\[S=\left(a + \frac{bn}2\right)(n+1).\]
