Question

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

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

Answer 1

`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`

Answer 2

`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`