How do you calculate #sum_(n=1)^(∞)(4/3)^(2-n)#?

How do you calculate #sum_(n=1)^(∞)(4/3)^(2-n)#?

2 Answers
Jan 24, 2018

#S=16/3#

Explanation:

.

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#

Jan 24, 2018

#sum_{n=1}^oo (4/3)^(2-n)=16/3#

Explanation:

#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#