1 + 3 + 9 + 27 + ... + 729?

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Answer 1 · 2018-01-09T14:23:13

#1093#

#1, 3, 9, 27...#
geometric progression
common ratio #r = 3#
starting term #a=1#
#u_n = 3^(n-1)#

sum of a geometric series: #a ((1-r^n)/(1-r))#

#r =3, a=1#

#a ((1-r^n)/(1-r)) = (1-r^n)/(1-r)#

#(1-r^n)/(1-r) = (1-(3^n))/(1-3)#

#=(1-(3^n))/(-2)#

when #u_n = 729#, #3^(n-1) = 729#
#3^(n-1) = 3^6 = 729#

#n-1 = 6#
#n=7#

when #u_n = 729#, #n=7#
(#729# is the #7#th term in the sequence)

#(1-(3^n))/(-2) = (1-3^7)/-2#

#=2186/2 = 1093#

#a ((1-r^n)/(1-r)) = 1093# when #n=7#, or when the sequence #u_n = 3^(n-1)# ends with #729#.