Reformulate an expression

Question

I want rewrite the product $\prod_{j=1}^k(a_j-b_j)$ into sums, i.e. multiplying this out.

For $k=2$ I have $(a_1-b_1)(a_2-b_2)=a_1a_2-a_1b_2-b_1a_2+b_1b_2$

For $k=3$ I have $(a_1-b_1)(a_2-b_2)(a_3-b_3)=a_1a_2a_3-a_1b_2a_3-b_1a_2a_3+b_1b_2a_3-a_1a_2b_3+a_1b_2b_3+b_1a_2b_3-b_1b_2b_3$

However, I failed to recognise the systematic and write this as $\sum..$

Answer 1

In principle one answer would be \begin{eqnarray*} \sum_{i_1=0}^{1} \cdots \sum_{i_k=0}^{1} \prod_{m=1}^{k} (a_m-(1-i_m)b_m). \end{eqnarray*} A better way would be to sum over the power set \begin{eqnarray*} \sum_{S \subseteq [m]} (-1)^{ \mid S \mid} X_S \end{eqnarray*} where $X_S$ is defined by the following product \begin{eqnarray*} \prod_{i \in S} b_i \prod_{i' \in [m] / S} a_{i'}. \end{eqnarray*}

Answer 2

$$ \prod_{j=1}^k (a_j - b_j) = \sum_{(s_1, \cdots, s_k) \in \{0,1\}^k}\left((-1)^{\sum_{j=1}^k s_j}\prod_{j=1}^k a_j^{1-s_j}b_j^{s_j}\right). $$