The entire sum can now be calculated. Fully simplify the expression. \\[ \\begin{array}{c} \\left(\\frac{b-a}{2 n^{2}}\ ight)\\left(\\sum_{k=1}^{n}((2 m b-2 m a) k)+\\sum_{k=1}^{n}(2 n m a-b m+a m+2 n c)\ ight) \\\\ \\downarrow \\\\ \\left(\\frac{b-a}{2 n^{2}}\ ight)\\left((2 m b-2 m a) \\frac{n(n+1)}{2}+2 n^{2} m a-b m n+a m n+2 n^{2} c\ ight) \\\\ \\downarrow \\\\ \\frac{b^{2} m+2 b c-a^{2} m-2 a c}{2} \\end{array} \\]