How do you find #\sum _ { n = 1} ^ { 9} ( \frac { 1} { 2} ) ^ { n - 1}#?

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Answer 1 · 2017-11-02T17:21:44

#511/256#

let's start by writing out the series and then identify what it is

#sum_(n=1)^9(1/2)^(n-1)=(1/2)^(1-1)+(1/2)^(2-1)+(1/2)^(3-1)+..(1/2)^(9-1)#

#=(1/2)^0+(1/2)^1+(1/2)^(2)+...(1/2)^8#

#=1+1/2+(1/2)^2+....+(1/2)^8#

we have a #GP#

#a=1#

#r=1/2#

#n=9#

for a GP

#S_n=(a(1-r^n))/(1-r), r!=1#

in this case

#S_n=(1(1-(1/2)^9))/(1-(1/2))#

#=2(1-(1/2)^9)#

#=2(1-1/512)=511/256#

note since

#|r|<1 #

the series has a sum to infinity which is

#S_oo=1/(1-(1/2))=2#