#log(2)+log(2/3)+log(2/3^2)+log(2/3^3)+...#. The sum of the first 10 terms equals?

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Answer 1 · 2018-01-11T15:27:49

#10*ln(2)-45ln(3)~~-42.5#

Use the logarithm rule #ln(a)+ln(b)=ln(ab)#

#f_1=ln(2)=ln(2/3^0)#

#f_2=ln(2)+ln(2/3)=ln(2^2/3^1)#

#f_3=ln(2)+ln(2/3)+ln(2/3^2)=ln(2^3/3^3)#

#f_4=ln(2)+ln(2/3)+ln(2/3^2)+ln(2/3^3)=ln(2^4/3^6)#

A pattern occurs

#f_n=ln(2^n/3^(((n^2-n)/2)))=nln(2)-(n^2-n)/2ln(3)#

For #n=10#

#f_10=10*ln(2)-(10^2-10)/2ln(3)#

#f_10=10*ln(2)-45ln(3)#

#f_10~~-42.5#