# How do you find the sum of Sigma (-2)^j where j is [0,4]?

Jul 7, 2017

${\sum}_{\text{j=0"^4}} {\left(- 2\right)}^{j} = 11$

#### Explanation:

${\sum}_{\text{j=0"^4}} {\left(- 2\right)}^{j} = {\left(- 2\right)}^{0} + {\left(- 2\right)}^{1} + {\left(- 2\right)}^{2} + {\left(- 2\right)}^{3} + {\left(- 2\right)}^{4}$

$= 1 - 2 + 4 - 8 + 16 = 11$

Jul 7, 2017

$11$

#### Explanation:

${\sum}_{r = 0}^{4} {\left(- 2\right)}^{j}$

$= {\left(- 2\right)}^{0} + {\left(- 2\right)}^{1} + {\left(- 2\right)}^{2} + {\left(- 2\right)}^{3} + {\left(- 2\right)}^{4}$

$= 1 - 2 + 4 - 8 + 16$

$= 11$