Consider the summation:
\[ \sum_{i=1}^n (a_i+b_i) = a_1 + b_1 + a_2 + b_2 + \cdots + a_{n-1} + b_{n-1} + a_n + b_n \]
Grouping all the \(a_i\)'s together and the \(b_i\)'s together:
\[\begin{align} \sum_{i=1}^n (a_i+b_i) &= (a_1 + a_2 + \cdots + a_{n-1} + a_n) + (b_1 + b_2 + \cdots + b_{n-1} + b_n) \\ &= \sum_{i=1}^n a_i+ \sum_{i=1}^n b_i \end{align}\]
Similar reasoning can be used to show that \(\sum_{i=1}^n (a_i - b_i) = \sum_{i=1}^n a_i - \sum_{i=1}^n b_i \).