Writing $a^2b^0c^0+a^0b^2c^0+a^0b^0c^2+a^1b^1c^0+a^1b^0c^1+a^0b^1c^1$ using $\sum$? $$a^2+b^2+c^2+ab+ac+bc$$
$$=a^2b^0c^0+a^0b^2c^0+a^0b^0c^2+a^1b^1c^0+a^1b^0c^1+a^0b^1c^1$$
Been messing around with some probability stuff and that popped up. I couldn't figure out how to write it in summation form so I can generalize it for when the sum of the exponents isn't just 2.

Let us call the elements $a_1,a_2$ and $a_3$ instead of $a,b$ and $c$. Then we can write $$\sum_{\substack{i_1,i_2,i_3\ge0\\ i_1+i_2+i_3=2}}\mkern-9mu a_1^{i_1} a_2^{i_2} a_3^{i_3}$$ for the sum that you are regarding. The reason to rename the elements is that you can also state it in terms of multi index notation then if you want to.
Edit: Let me elaborate on the latter.  If we write $a = (a_1,a_2,a_3)$ and $i = (i_1,i_2,i_3)$ and consider $a^i$ as $a_1^{i_1} a_2^{i_2} a_3^{i_3}$ (so we work entrywise), then we can express the sum as $$\sum_{i \in \mathbb{N}_2^3} a^i,$$ where $\mathbb{N}_2^3 = \lbrace (i_1,i_2,i_3) \in \mathbb{N}^3 \mid i_1 +i_2 +i_3 = 2 \rbrace$.
