7.9: Sums of Finite Geometric Series
Anna is on a progressive workout plan, every day she adds 5% to her exercise time. If she starts by exercising 15mins on the first day, how many minutes will she have exercised all together on day 45?
This is a geometric series, since the difference between the exercise time on any two days is greater than the difference between any prior two days. You could just add:
Watch This
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 Khan Academy: Geometric series sum to figure out mortgage payments
Guidance
A finite geometric series is simply a geometric series with a specific number of terms. For example, consider the series: 50 + 25 + 12.5 + ....The series is geometric: the first term is 50, and the common ration is (1/2).
 The sum of the first two terms is 50 + 25 = 75. We can write this as S_{2} = 75
 The sum of the first three is 50 + 25 + 12.5 = 87.5. We can write this as S_{3} = 87.5
To find the value of S_{n} in general, we could simply add together the first n terms in a series. However, this would obviously be tedious for a large value of n. Given the regular pattern in a geometric series  every term is (1/r) of the previous term, and the n^{th} term is a_{n} = a_{1}r^{n  1} , we can use induction to prove a formula for S_{n} .
The sum of the first n terms in a geometric series is
For example, for the series 50 + 25 + 12.5 + ... , the sum of the first 6 terms is:
The figure below shows the same calculation on a TI83/4 calculator:
We can use this formula as long as the series in question is geometric.
Example A
Find the sum of the first 10 terms of a geometric series with a_{1} = 3 and r = 5.
Solution
The sum is 58,593.

Sn=a1(1−rn)1−r=3(1−57)1−5=3(1−78,125)−4=3(−78,124)−4=58,593  Notice that because the common ratio in this series is 5, the terms get larger and larger. This means that for increasing values of n the sums will also get larger and larger. In contrast, in the series with common ratio (1/2), the terms gets smaller and smaller. This situation implies something important about the sum.
Example B
Find the sum of each series:
 a) The first term of a geometric series is 4, and the common ratio is 3. Find S_{8}.
 b) The first term of a geometric series is 80, and the common ratio is (1/4). Find S_{7}.
Solution

a)
S8=4(1−38)1−3=13,120 
b)
S7=80(1−(14)7)1−14≈106.66
Example C
Prove the formula
Solution

1. If n = 1, the n^{th} sum is the first sum, or a_{1} . Using the hypothesized equation, we get
S1=a1(1−r1)1−r=a1(1−r)1−r=a1 . This establishes the base case. 
2. Assume that the sum of the first k terms in a geometric series is
Sk=a1(1−rk)1−r . 
3. Show that the sum of the first k+1 terms in a geometric series is
Sk+1=a1(1−rk+1)1−r . 

Sk+1=Sk+ak+1 The k+1 sum is thekth sum, plus thek+1 term=a1(1−rk)1−r+a1rk+1−1 Substitute from step 2, and substitute the k+1 term=a1(1−rk)1−r+a1rk(1−r)1−r The common denominator is 1−r =a1(1−rk)+a1rk(1−r)1−r Simplify the fraction =a1[1−rk+rk(1−r)]1−r =a1[1−rk+rk−rk+1]1−r =a1[1−rk+1]1−r It is proven.

Therefore we have shown that
Concept question wrapup "Anna is on a progressive workout plan, every day she adds 5% to her exercise time. If she starts by exercising 15mins on the first day, how many minutes will she have exercised all together on day 45?"
Use the formula:


Vocabulary
A finite series has a defined ending value.
An infinte series does not have a defined ending value.
A geometric series is a series where the difference between terms increases or decreases between each pair of terms.
Guided Practice
Questions
1) Find the sum:
2) Use a geometric series to answer the question:
 In January, a company’s sales totaled $11,000. It is predicted that the company’s sales will increase 5% each month for the next year. At this rate, what will be the total sales for the year?
3) Write the first 5 terms of the sequence:
4) Write the 3rd, 4th, and 6th terms of:
5) Find the sum of the series:
Solutions
1)
2)
3) Just do the multiplication for each term

−5⋅340 to−5⋅1→−5 ..... forn=0 
−5⋅341 to−5⋅34→−154→−3.75 ..... forn=1 
−5⋅342 to−5⋅916→−4516→−2.8 ..... forn=2 
−5⋅343 to−5⋅2764→−13564→−2.1 ..... forn=3 
−5⋅344 to−5⋅81256→−405256→−1.6 ..... forn=4

∴ the first 5 terms are:−5,−3.75,−2.8,−2.1,−1.6
4) As with problem 3, just perform the operations on the indicated values of n:

332→33−−√→27−−√→5.2 ..... forn=3 
342→9 ..... forn=4 
362→33→27 ..... for \begin{align*}n = 6\end{align*}
 \begin{align*}\therefore\end{align*} the 3rd, 4th, and 6th terms are: \begin{align*}5.2, 9, 27\end{align*}
5) To find the sum of the series \begin{align*}\sum_{n = 1}^6 \left(\frac{3}{2}\right)^{n1}\end{align*}

We could calculate all of the values for \begin{align*}n = 1 \to 6\end{align*} and add them, getting:
 \begin{align*}1+ \frac{3}{2} + \frac{9}{4} + \frac{27}{8} + \frac{81}{16} + \frac{243}{32} = \frac{133}{32}\end{align*}

Or we can use the formula: \begin{align*}\left( \frac{1  r^k}{1  r} \right)\end{align*}
 \begin{align*}\left( \frac{1  \left(\frac{3}{2}\right)^6}{1  \left(\frac{3}{2}\right)} \right) = \frac{133}{32}\end{align*}
Practice
Find the sum of the finite series. You may simply calculate the individual terms and add them, or you may use the formula: \begin{align*}S_n = \frac{a_1 (1 r^n)}{1 r}\end{align*}
 \begin{align*}1 + \left(\frac{1}{2}\right) + \frac{1}{4} + ... + \frac{1}{64}\end{align*}
 \begin{align*}6 + 12 24 + ... 6144\end{align*}
 \begin{align*}(4) + (12) + (36) + ... + (2916)\end{align*}
 \begin{align*}9 + (45) + 225 + ... + (17,578,125)\end{align*}
 \begin{align*}\sum_{n=1}^5 (2)^{n1}\end{align*}
 \begin{align*}\sum_{n=1}^{10} 6\left(\frac{1}{2}\right)^{n1}\end{align*}
 \begin{align*}\sum_{n=1}^6 8 \cdot 3^{n1}\end{align*}
 \begin{align*}(5) + 10 + (20) + ... + (1280)\end{align*}
 \begin{align*}(9) + \frac{9}{2} + \left(\frac{9}{4}\right) + ... + \left(\frac{9}{16}\right)\end{align*}
 \begin{align*}\sum_{n=1}^9 4 \cdot \left(\frac{1}{2}\right)^{n1}\end{align*}
 \begin{align*}\sum_{n=1}^{10} 4 \cdot \left(\frac{2}{3}\right)^{n1}\end{align*}
 \begin{align*}(3) + \left(\frac{3}{2}\right) + \left(\frac{3}{4}\right) + ... + \left(\frac{3}{1024}\right)\end{align*}
 \begin{align*}\sum_{n=1}^{11} 9 \cdot (2)^{(n1)}\end{align*}
 \begin{align*}\sum_{n=1}^9 9 \cdot \left(\frac{5}{3}\right)^{n1}\end{align*}
 \begin{align*}\sum_{n=1}^6 2 \cdot \left(\frac{5}{4}\right)^{n1}\end{align*}
 \begin{align*}\sum_{n=1}^{11} 8 \cdot \left(\frac{1}{3}\right)^{n1}\end{align*}
finite series
A series is finite if it has a defined ending value.geometric series
A geometric series is a geometric sequence written as an uncalculated sum of terms.induction
Induction is a method of mathematical proof typically used to establish that a given statement is true for all positive integers.infinite series
An infinite series is the sum of the terms in a sequence that has an infinite number of terms.series
A series is the sum of the terms of a sequence.Image Attributions
Description
Learning Objectives
Here you will learn how to calculate the sum of a series when the difference between the members is not constant.