TITLE: Calculating double sum $\sum_{i=-1}^{5}\sum_{j = -5}^{1}(1 + 2i + 3j + 4ij)$
QUESTION [6 upvotes]: How can I calculate sums of the form:
$$\sum_{i=-1}^{5}\sum_{j = -5}^{1}(1 + 2i + 3j + 4ij)?$$
Shall I calculate it like first for $i = -1, j = \{-5,\dots, 1\}$, then for $i = 2, j = \{-5, \dots, 1\}$? Is there some shortcut for calculating those kinds of sums, i.e. something akin to summation of arithmetic series?

REPLY [12 votes]: If you write
$$1+2i+3j+4i j=\frac{1}{2}(4i+3)(2j+1)-\frac{1}{2}$$
you can use the sum of arithmetic sequence formula
$$\frac{1}{2}\sum _{i=-1}^5 (4i+3)\sum _{j=-5}^1 (2j+1)-\frac{1}{2}49$$
$$=\frac{1}{2}(77)(-21)-\frac{1}{2}49$$
$$=\text{-833}$$