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

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Answer 1 · 2017-07-07T11:32:58

#sum_"j=0"^4"(-2)^j = 11#

#sum_"j=0"^4"(-2)^j = (-2)^0 + (-2)^1 +(-2)^2 + (-2)^3 +(-2)^4#

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

Answer 2 · 2017-07-07T12:47:08

#11#

#sum_(r=0)^4(-2)^j#

#=(-2)^0+(-2)^1+(-2)^2+(-2)^3+(-2)^4#

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

#=11#