Answer
$$\sum\limits_{n = 1}^6 {\left( {2n - 1} \right)} {\left( { - 1} \right)^{n + 1}}$$
Work Step by Step
$$\eqalign{
& 1 - 3 + 5 - 7 + 9 - 11 \cr
& d = 3 - 1 = 5 - 3 = 7 - 5 = 2,{\text{ }}{a_1} = 1,{\text{ }}n = 15 \cr
& {a_n} = {a_1} + \left( {n - 1} \right)d \cr
& {a_n} = 1 + \left( {n - 1} \right)\left( 2 \right) \cr
& {a_n} = 1 + 2n - 2 \cr
& {a_n} = 2n - 1 \cr
& {\text{The sign of the terms are alternating, then we must add }}{\left( { - 1} \right)^n} \cr
& \sum\limits_{n = 1}^6 {\left( {2n - 1} \right)} {\left( { - 1} \right)^{n + 1}} \cr} $$
