Answer
$\sum\limits_{i =1}^{n}c=nc$
Work Step by Step
Theorem 3(b) defines as
$\sum\limits_{i =1}^{n}c=nc$
The left side can be written as
$\sum\limits_{i =1}^{n}c=c+c+c+c+c....+c$ (n- times)
Since $c$ has been added n-times, it can be generally multiplied by $n$ as $nc$.
Hence, $\sum\limits_{i =1}^{n}c=nc$
