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

$11$

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

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

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

$=11$

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