Rules for using multi-indices

Question

I've only encountered multi-index notation in the context of the multinomial theorem. There, the notation is used like this: $(k_1 + k_2 +...+k_m) = n, \ k_i \in \Bbb N_{\ge 0}^n$, iterate over every permutation of the summation.

So basically: $$\sum_{(k_1 + k_2 + k_3) = 2} k = (1+1+0)+(1+0+1) + (0+1+1) +(2+0+0)+(0+2+0)+(0+0+2)$$

Is this true, or are there other rules?

Answer 1

The left-hand has to be written somewhat differently. We can write \begin{align*} \sum_{{k_1 + k_2 + k_3 = 2}\atop{k_1,k_2,k_3\geq 0}} \left(k_1+k_2+k_3\right) &= (1+1+0)+(1+0+1) + (0+1+1)\\ &\qquad+(2+0+0)+(0+2+0)+(0+0+2)\tag{1} \end{align*}

On the other hand OPs left-hand expression can be written as \begin{align*} \sum_{{k_1 + k_2 + k_3 = 2}\atop{k_1,k_2,k_3\geq 0}}k &=k\sum_{{k_1 + k_2 + k_3 = 2}\atop{k_1,k_2,k_3\geq 0}}1\tag{2}\\ &=6k\tag{3} \end{align*}

Comment:

• In (2) we note that $k$ is a constant which does not depend on any of the indices $k_1,k_2,k_3$. We can therefore factor out $k$ leaving a term $1$ within the scope of the sum.

• In (3) we recall that the index range specifies $6$ valid triples $(k_1,k_2,k_3)$ with $k_1+k_2+k_3=2$. This is explicitly shown at the right-hand side of (1). Since there are $6$ valid terms, we can simplify the sum to $6$.