## Calculus: Early Transcendentals 8th Edition

$\sum\limits_{i =1}^{n}c=nc$
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, $\sum\limits_{i =1}^{n}c=nc$