Answer
$$\sum\limits_{i = 1}^5 {\left( {4i - 1} \right)} = 55$$
Work Step by Step
$$\eqalign{
& \sum\limits_{i = 1}^5 {\left( {4i - 1} \right)} \cr
& {\text{The first term is}} \cr
& {a_1} = 4\left( 1 \right) - 1 = 3 \cr
& {\text{The last term is}} \cr
& {a_5} = 4\left( 5 \right) - 1 = 19 \cr
& {\text{Using the formula }}{S_n} = \frac{n}{2}\left( {{a_1} + {a_n}} \right),{\text{ we obtain}} \cr
& \sum\limits_{i = 1}^5 {\left( {4i - 1} \right)} = {S_5} = \frac{5}{2}\left( {3 + 19} \right) \cr
& \sum\limits_{i = 1}^5 {\left( {4i - 1} \right)} = 55 \cr} $$
