What is #sum_(n=1)^(4) [100(-4)^(n-1)]#?

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2 Answers
Narad T.
May 22, 2018

The answer is #=-5100#

Explanation:

This is a geometric progression

#r=-4#

#a_n=100((-4)^(n-1))#

#a_(n+1)=100((-4)^(n+1-1))#

#a_1=100#

The sum is

#S=a_1(1-r^(n+1))/(1-r)#

#=100(1-(-4)^4)/(1-(-4))#

#=100/5(1-256)#

#=-5100#

Truong-Son N.
Jul 1, 2018

Here is an alternative, simple approach. Since there are only 4 terms in the sum, just expand it.

#sum_(n=1)^(4) [100(-4)^(n-1)] = ?#

There is no #n# dependence in #100#, so factor it out:

#= 100 sum_(n=1)^(4) (-4)^(n-1)#

Now expand:

#= 100[(-4)^(1-1) + (-4)^(2-1) + (-4)^(3-1) + (-4)^(4-1)]#

#= 100[(-4)^(0) + (-4)^(1) + (-4)^(2) + (-4)^(3)]#

#= 100(1 - 4 + 16 - 64)#

#= 100(-51)#

#= color(blue)(-5100)#

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