## Algebra 2 (1st Edition)

$\displaystyle \sum_{i=1}^{n}ka_{i}=ka_{1}+ka_{2}+ka_{3}+ka_{4}+\cdots+ka_{n}$ ... we can factor k out, $=k(a_{1}+a_{2}+a_{3}+\cdots+a_{n})$ ... and recognize the sum in the parentheses can be written in summation notation, The lower limit is 1, and the upper limit is n. $=k\displaystyle \cdot\sum_{i=1}^{n}a_{i}$ The statement is true.