How do you find the sum of the geometric series #Sigma 81 (1/3)^(n-1)# n=1 to 7?

1 Answer
Mar 12, 2017

Desired sum is #121 4/9#

Explanation:

#Sigma81(1/3)^(n-1)# represents a geometric series whose first term is #a_1=81# and common ratio #r# is #1/3#. This can be easily found by putting values of #n# from #1# onward and series appears as

#81,27,9,3,1,....................#

The sum of such series up to first #n# terms is given by

#S_n=(81(1-r^n))/((1-r))#

Hence, in the given series the desired sum is

#(81(1-(1/3)^7))/((1-1/3))#

= #81(1-1/3^7)xx3/2#

= #81(2186/2187)xx3/2#

= #cancel81(cancel(2186)^1093/cancel(2187)^27)xx3/cancel2#

= #1093/9=121 4/9#