How do you calculate #sum_(n=1)^(∞)(4/3)^(2-n)#?
How do you calculate #sum_(n=1)^(∞)(4/3)^(2-n)#?
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First term #(n=1)#, is:
#(4/3)^(2-1)=(4/3)^1=4/3#
Second Term (#n=2#), is:
#(4/3)^(2-2)=(4/3)^0=1#
Third term #(n=3)#, is:
#(4/3)^(2-3)=(4/3)^(-1)=1/(4/3)^1=1/(4/3)=3/4#
Fourth Term #(n=4)#, is:
#(4/3)^(2-4)=(4/3)^(-2)=1/(4/3)^2=1/(16/9)=9/16#
Fifth term #(n=5)#, is:
#(4/3)^(2-5)=(4/3)^(-3)=1/(4/3)^3=1/(64/27)=27/64#
As can be seen, the first five terms of this series are:
#4/3, 1, 3/4, 9/16, 27/64#
This is a geometric series with a common ratio of #(a_(n+1)/a_n)=a_2/a_1=1/(4/3)=3/4#:
#r=3/4#
If #absr < 1# there is a formula for the sum of infinite geometric series:
#S=a_1/(1-r)#
#S=(4/3)/(1-(3/4))=(4/3)/(1/4)=4/3*4/1=16/3#
#sum_{n=1}^oo (4/3)^(2-n)=16/3#
#sum_{n=1}^oo (4/3)^(2-n)#
=#sum_{n=1}^oo (3/4)^(n-2)#
=#sum_{n=0}^oo (3/4)^(n-1)#
=#(3/4)^(-1)*1/(1-3/4)#
=#(4/3)/(1/4)#
=#16/3#
