Calculating double sum $\sum_{i=-1}^{5}\sum_{j = -5}^{1}(1 + 2i + 3j + 4ij)$ 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?
 A: One way ,you can start from $i$,and in this situation $j$  should be considered as constant $$\sum _{ i=-1 }^{ 5 } \sum _{ j=-5 }^{ 1 } (1+2i+3j+4ij)=\sum _{ i=-1 }^{ 5 } \left( \sum _{ j=-5 }^{ 1 } +2i\sum _{ j=-5 }^{ 1 } +3\sum _{ j=-5 }^{ 1 } j+4i\sum _{ j=-5 }^{ 1 }  \right) =\\ =\sum _{ i=-1 }^{ 5 } \left( 7+2i\cdot 7+3\left( -5-4-3-2-1+0+1 \right) +4i\cdot 7 \right) =\\=\sum _{ i=-1 }^{ 5 } \left( 42i-35 \right) =42\sum _{ i=-1 }^{ 5 } i-35\sum _{ i=-1 }^{ 5 } =\\ =42\left( -1+0+1+2+3+4+5 \right) -35\cdot 7=-588-245=-833$$
A: The method that you propose is certainly the most straightforward, but it is a bit tedious. You can speed things up by breaking it up into four sums and dealing with each of them separately:
$$\sum_{i=-1}^5\sum_{j=-5}^11+2\sum_{i=-1}^5\sum_{j=-5}^1i+3\sum_{i=-1}^5\sum_{j=-5}^1j+4\sum_{i=-1}^5\sum_{j=-5}^1ij\;.$$
The first one is easy: you add once for each possible pair $\langle i,j\rangle$, and there are $7^2=49$ possible pairs, so the total is $49$. (There are $7$ values of $i$ and $7$ values of $j$.)
The second one takes just a little more work. For each value of $i$ in the outer sum you’re simply adding up $7$ copies of $i$ in the inner sum, so
$$\sum_{i=-1}^5\sum_{j=-5}^1i=7\sum_{i=-1}^5i\;,$$
and that last summation is just an arithmetic series.
For the third one you can sum the arithmetic series $\sum_{j=-1}^1j$ and then observe that the outer sum is just adding up $7$ copies of this total.
That leaves the most interesting one. Notice that $i$ is a constant in the inner summation: it doesn’t depend on $j$. Thus,
$$\sum_{i=-1}^5\sum_{j=-5}^1ij=\sum_{i=-1}^5i\sum_{j=-5}^1j\;.$$
The inner summation is now just the same arithmetic series as in the third summation, so you can evaluate it and pull out the sum as a constant multiplier. All that remains then is to sum the arithmetic series $\sum_{i=-1}^5i$ and multiply the result by that multiplier (and by $4$, of course).
A: 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
$$\frac{1}{2}\sum _{i=-1}^5 (4i+3)\sum _{j=-5}^1 (2j+1)-\frac{1}{2}49$$
$$=\frac{1}{2}(77)(-21)-\frac{1}{2}49$$
$$=\text{-833}$$
A: Rewrite $1+2i+3j+4ij$ as $(1+2i)(1+2j)+(1)(j)$. Then the double sums of products of two factors that depends only on one index can be factored ($\sum\sum ab=\sum a\sum b$):
$$\sum_{i=-1}^{5}\sum_{j = -5}^{1}(1 + 2i + 3j + 4ij)=\sum_{i=-1}^{5}(1+2i)\sum_{j = -5}^{1}(1+2j)+\sum_{i=-1}^{5}1\sum_{j = -5}^{1}j\\
=35(-21)+7(-14)=-833.$$

To speed-up computation, you can notice that
$$\sum_{-1}^5 i=\sum_{2}^5 i=2+3+4+5=14,$$
then 
$$\sum_{-1}^5(1+2i)=7+2\cdot14=35,\\
\sum_{-5}^1 j=-14,\\
\sum_{-5}^1(1+2j)=7-2\cdot14=-21.$$
