How do you find the fifth partial sum of #sum_(i=1)^oo 2(1/3)^i#?

1 Answer
Jul 2, 2017

Answer:

#sum_(i=1)^5 2(1/3)^i = 242/243#

Explanation:

This is a geometric series with initial term #a=2/3# and common ratio #r = 1/3#.

The general term of the series may be written:

#a_i = ar^(i-1)#

We find:

#(1-r) sum_(i=1)^n a_i = sum_(i=1)^n ar^(i-1) - r sum_(i=1)^n ar^(i-1)#

#color(white)((1-r) sum_(i=1)^n a_i) = sum_(i=1)^n ar^(i-1) - sum_(i=2)^(n+1) ar^(i-1)#

#color(white)((1-r) sum_(i=1)^n a_i) = a+color(red)(cancel(color(black)(sum_(i=2)^n ar^(i-1)))) - color(red)(cancel(color(black)(sum_(i=2)^n ar^(i-1)))) - ar^n#

#color(white)((1-r) sum_(i=1)^n a_i) = a(1-r^n)#

So dividing both ends by #(1-r)#, we find:

#sum_(i=1)^n a_i = (a(1-r^n))/(1-r)#

So in the given example, with #a=2/3# and #r=1/3#, we find:

#sum_(i=1)^5 2(1/3)^i = (2/3(1-(1/3)^5))/(1-(1/3))#

#color(white)(sum_(i=1)^5 2(1/3)^i) = (2/3(1-1/243))/(2/3)#

#color(white)(sum_(i=1)^5 2(1/3)^i) = 242/243#