How do you find the fourth partial sum of #Sigma 5(1/2)^i# from i=1 to #oo#?

1 Answer
Jul 12, 2018

Answer:

#75/16#

Explanation:

A series is an infinite sum, i.e. a sum with infinite terms.

Partial sums are approximations of the series, where you reduce it to a finite number of terms. In fact, the #n^"th"# partial sum is the sum of the first #n# terms, and the series is the limit of the partial sums as #n \to \infty#. You can imagine this, for example, as approaching the infinite sum by summing the first #10#, #100#, #1000#, #100000#, #1000000000#,... terms.

So, the fourth partial sum is the sum of the first #4# terms. Since #i# starts from one, we must sum the terms corresponding to #i=1,2,3,4#.

First of all, let's factor the #5# out of the sum:

#\sum_{i=1}^4 5 (1/2)^i = 5\sum_{i=1}^4 (1/2)^i#

Now we can simply translate the notations, plugging #1,2,3,4# in place of #i# for each term:

#5\sum_{i=1}^4 (1/2)^i = 5(1/2+1/4+1/8+1/16)5*15/16=75/16#