How do you find sum of geometric series 2+6+18+...+1458?

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Answer 1 · 2016-03-15T08:02:03

#2186#

Given that first term #a_1=2#

Common ratio #q=a_2/a_1=6/2=3#

General expression for #nth# term is

#a_n=a_1*q^(n-1)#

To ascertain what is number of the last term, insert in the above equation

#1458=2*3^(n-1) #
#=> 3^(n-1) = 1458/2=729 #
Writing #729# as power of #3# we obtain

# 3^(n-1) = 3^6#, comparing the exponents
# n-1=6#
or #n=7#
Now sum of #n# terms is given by the expression

#S_n = (a_1(1-q^n))/(1-q)#

#S_7 = (2 (1-3^7)) / (1-3) #
#= (2 (1-2187))/-2 = 2186#