Answer
$$\sum\limits_{n = 1}^8 {\left( {2n - 1} \right)} $$
Work Step by Step
$$\eqalign{
& 1 + 3 + 5 + 7 + \cdots + 15 \cr
& d = 3 - 1 = 5 - 3 = 7 - 5 = 2,{\text{ }}{a_1} = 1,{\text{ }}n = 15 \cr
& {a_n} = {a_1} + \left( {n - 1} \right)d \cr
& {a_n} = 1 + \left( {n - 1} \right)\left( 2 \right) \cr
& {a_n} = 1 + 2n - 2 \cr
& {a_n} = 2n - 1 \cr
& {\text{Then,}} \cr
& \sum\limits_{n = 1}^8 {\left( {2n - 1} \right)} \cr} $$
