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.
 A: $$\sum_{\substack{i,j,k\ge0\\ i+j+k=2}}\mkern-9mua^i b^j c^k$$
seems to be what you're after.
A: 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 $\mathbb{N}_2^3 = \lbrace (i_1,i_2,i_3) \in \mathbb{N}^3 \mid i_1 +i_2 +i_3 = 2 \rbrace$.
A: For example : try $$\sum \{ a^ib^jc^k: i,j,k \in \Bbb N, i+j+k=2\}$$
But what does writing such an expression this way buy you?
A: Consider
$$\frac12\left(\sum_i x_i\right)^2+\frac12\sum_i x_i^2.$$
This will not generalize easily, though.
