How to prove that $(\sum_{i=1}^n a_i)(\sum_{i=1}^n b_i)= \sum_{i,j} a_ib_j$?

Question

How to prove that $(\sum_{i=1}^n a_i)(\sum_{i=1}^n b_i)= \sum_{i,j} a_ib_j$? Is there any way to visualize the sums on both sides.

Answer 1

Assuming $a_i \ge 0$ and $b_j \ge 0$, draw a rectangle with side lengths $\sum_i a_i$ and $\sum_j b_j$, and partition it into $n^2$ rectangles with side lengths $a_i$ and $b_j$. Now compute the total area in two different ways.

Answer 2

$$\sum_{i,j} a_ib_j =\sum_{i\in\{1,\ldots,n\},\ j\in\{1,\ldots,n\}} a_ib_j = \sum_{j=1}^n a_1 b_j+ \sum_{j=1}^n a_2 b_j + \ldots + \sum_{j=1}^n a_n b_j = \left(\sum_{i=1}^n a_i\right)\left(\sum_{i=1}^n b_i\right).$$