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

Answer 1

Charles Anctil

#11#

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

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

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

#=11#

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Answer 2

Samantha Adkison

#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#

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Answer 3

Scarlett Allison

To find the sum of ( \sum_{j=0}^{4} (-2)^j ), you substitute the values of ( j ) from 0 to 4 into the expression ( (-2)^j ) and then add them together.

So, ( j = 0: (-2)^0 = 1 )
( j = 1: (-2)^1 = -2 )
( j = 2: (-2)^2 = 4 )
( j = 3: (-2)^3 = -8 )
( j = 4: (-2)^4 = 16 )

Adding these values together:
( 1 + (-2) + 4 + (-8) + 16 = 11 )

Therefore, the sum of ( \sum_{j=0}^{4} (-2)^j ) is 11.

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.