How do you find #S_n# for the geometric series #a_1=162#, r=1/3, n=6?

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Answer 1 · 2016-12-01T07:11:25

The answer is #=2186/9#

Let #S_n# be the sum of a geometric series

#S_n=a_1+a_1r+ ........ a_1r^n#

Then, #rS_n=a_1r+..................+a_1r^(n+1)#

Therefore,

#S_n(1-r)=a_1(1-r^(n+1))#

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

#a_1=162#

#r=1/3#

#n=6#

#S_6=162(1-(1/3)^7)/(1-1/3)#

#=162*3/2(1-(1/3)^7)#

#=243-243/3^7#

#=243-1/9#

#=2186/9#