How do you find the sum of the infinite geometric series given #Sigma 48(2/3)^(n-1)# from n=1 to #oo#?

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Answer 1 · 2016-10-25T10:43:59

#144#

First write out the GP, taking the #48# out as a common factor ( to make the calculations easier)

#48sum_1^oo(2/3)^(n-1)=48(1+2/3+(2/3)^2+(2/3)^3+.....+)#

for this is GP first term #a=1#, common ratio # r=2/3#

the sum to infinity of GP is #S_(OO)=a/(1-r)#

providing # |r|<1#

in this case # r=2/3<1, :> S_(OO)# exists

#S_(OO)=a/(1-r)=48(1/(1-(2/3)))#

#=48(1/(1/3))=48xx3=144#