How do you find the sum of the geometric series #Sigma 2(-3)^(n-1)# n=1 to 6?

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Answer 1 · 2017-02-06T12:00:20

#= - 364#

The series is: #S = Sigma_(n = 1)^(6) 2(-3)^(n-1) = 2 Sigma_(n = 1)^(6) (-3)^(n-1)#

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

Which can be written as:

#S = 2( 1 + r + .... + r^5) #, with #r = -3#

#implies r S = 2( r + r^2 + .... + r^6) #

#implies rS - S = (r-1) S = 2( r^6 -1) #

#implies S = 2( r^6 -1)/(r-1) #

#= 2( (-3)^6 -1)/((-3)-1) = - 364#