Use of Multinomial theorem.

Question

I have the next identity which I want to prove.

$$(\sum_{j}k_j^2)^{s} = \sum_{b_1+\ldots+ b_n =s} \prod_j k_j^{2b_j}$$

Obviously I need to use the Multinomial theorem, but how to procceed from there?

$$(\sum_{j}k_j^2)^{s} = \prod_{j=1}^n \sum_{\sum_i m_i^j =b_j} {b_j\choose m_1^j,\ldots , m_n^j} \prod_{1\leq t \leq n}k_t^{2m_t^j}$$

Thanks in advance. PS $s$ is a non-negative integer s.t $s=b_1+\ldots b_n $.

Answer 1

This isn't the multinomial theorem; this is just the distributive law. If, for each $i=1,\cdots,s$ we have a set $\{a_{i,j_i}\mid j_i \in J_i\}$, then

$$\prod_{i=1}^s\sum_{j_i \in J_i} a_{i,j_i} = \sum_{j_1,\cdots,j_s}\prod_{i=1}^s a_{i,j_i}.$$

In your case the sets are all the same $(a_{i,j_i} = a_j = k_j^2)$, so each of the products on the right can be expressed as a product of powers.