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: \begin{align*}15 + (15 \cdot 1.05) + [(15 \cdot 1.05) \cdot 1.05]...\end{align*}
Watch This
Embedded Video:
 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 \begin{align*}S_n = \frac{a_1(1  r^n)} {1  r}\end{align*}
For example, for the series 50 + 25 + 12.5 + ... , the sum of the first 6 terms is:
\begin{align*}S_n = \frac{a_1(1  r^n)} {1  r} = \frac{50 \left(1  (\frac{1} {2})^6 \right)} {1  \frac{1} {2}} = \frac{50 \left(1  \frac{1} {64} \right)} {\frac{1} {2}} = \frac{50 \left(\frac{63} {64} \right)} {\frac{1} {2}} = 50 \left(\frac{63} {64} \right) \left(\frac{2} {1} \right) = 98 \frac{7} {16}\end{align*}
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.
 \begin{align*}S_n = \frac{a_1(1  r^n)} {1  r} = \frac{3(1  5^7)} {1  5} = \frac{3(1  78,125)} {4} = \frac{3(78,124)} {4} = 58,593\end{align*}
 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) \begin{align*}S_{8}=\frac {4(13^8)}{13}=13,120\end{align*}
 b) \begin{align*}S_{7}= \frac {80 \left ( {1 \left ( \frac{1}{4} \right )^7} \right )}{1\frac {1}{4}} \approx 106.66 \end{align*}
Example C
Prove the formula \begin{align*}S_{n}=\frac{a_1(1  r^n)} {1  r}\end{align*} by induction
Solution
 1. If n = 1, the n^{th} sum is the first sum, or a_{1} . Using the hypothesized equation, we get \begin{align*}S_1 = \frac{a_1(1  r^1)} {1  r} = \frac{a_1(1  r)} {1  r} = a_1\end{align*}. This establishes the base case.
 2. Assume that the sum of the first k terms in a geometric series is \begin{align*}S_k = \frac{a_1(1  r^k)} {1  r}\end{align*}.
 3. Show that the sum of the first k+1 terms in a geometric series is \begin{align*}S_{k + 1} = \frac{a_1(1  r^{k + 1})} {1  r}\end{align*}.


\begin{align*}S_{k + 1} = S_k + a_{k + 1}\end{align*} The \begin{align*}k + 1\end{align*} sum is the \begin{align*}k^{th}\end{align*} sum, plus the \begin{align*}k + 1\end{align*} term \begin{align*} = \frac{a_1(1  r^k)} {1  r} + a_1r^{k + 1  1}\end{align*} Substitute from step 2, and substitute the \begin{align*}k + 1\end{align*} term \begin{align*}= \frac{a_1(1  r^k)} {1  r} + \frac{a_1r^k(1  r)} {1 r}\end{align*} The common denominator is \begin{align*}1  r\end{align*} \begin{align*}= \frac{a_1(1  r^k) + a_1r^k(1  r)} {1  r}\end{align*} Simplify the fraction \begin{align*}= \frac{a_1 \left[1  r^k + r^k(1  r) \right]} {1  r}\end{align*} \begin{align*}= \frac{a_1 \left[1  r^k + r^k  r^{k + 1} \right]} {1  r}\end{align*} \begin{align*}= \frac{a_1 \left[1  r^{k + 1} \right]} {1  r}\end{align*} It is proven.

Therefore we have shown that \begin{align*}S_{n}=\frac{a_1(1  r^n)} {1  r}\end{align*} for a geometric series. Now we can use this equation to find any sum of a geometric series.
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: \begin{align*}S_{n}=\frac{a_1(1  r^n)} {1  r}\end{align*}


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: \begin{align*}5 + 10 + 20 +...+ 640\end{align*} (Hint: if a_{n} = 640 , what is n?)
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: \begin{align*}5 \cdot \frac{3}{4}^n\end{align*}
4) Write the 3rd, 4th, and 6th terms of: \begin{align*}(3)^{\left(\frac{N}{2}\right)}\end{align*}
5) Find the sum of the series: \begin{align*}\sum_{n = 1}^6 \left(\frac{3}{2}\right)^{n1}\end{align*}
Solutions
1) \begin{align*}S_{8}=\frac {5(12^8)}{12} = 1275\end{align*}
2) \begin{align*}S_{12}=\frac {11000(11.05^{12})}{11.05} = $175,088.39\end{align*}
3) Just do the multiplication for each term \begin{align*}n = 0 \to n = 4\end{align*}
 \begin{align*}5 \cdot \frac{3}{4}^0 \ to 5 \cdot 1 \to 5\end{align*} ..... for \begin{align*}n = 0\end{align*}
 \begin{align*}5 \cdot \frac{3}{4}^1 \ to 5 \cdot \frac{3}{4} \to \frac{15}{4} \to 3.75\end{align*} ..... for \begin{align*}n = 1\end{align*}
 \begin{align*}5 \cdot \frac{3}{4}^2 \ to 5 \cdot \frac{9}{16} \to \frac{45}{16} \to 2.8\end{align*} ..... for \begin{align*}n = 2\end{align*}
 \begin{align*}5 \cdot \frac{3}{4}^3 \ to 5 \cdot \frac{27}{64} \to \frac{135}{64} \to 2.1\end{align*} ..... for \begin{align*}n = 3\end{align*}
 \begin{align*}5 \cdot \frac{3}{4}^4 \ to 5 \cdot \frac{81}{256} \to \frac{405}{256} \to 1.6\end{align*} ..... for \begin{align*}n = 4\end{align*}
 \begin{align*}\therefore\end{align*} the first 5 terms are: \begin{align*}5, 3.75, 2.8, 2.1, 1.6\end{align*}
4) As with problem 3, just perform the operations on the indicated values of n:
 \begin{align*}3^{\frac{3}{2}} \to \sqrt{3^3} \to \sqrt{27} \to 5.2\end{align*} ..... for \begin{align*}n = 3\end{align*}
 \begin{align*}3^{\frac{4}{2}} \to 9\end{align*} ..... for \begin{align*}n = 4\end{align*}
 \begin{align*}3^{\frac{6}{2}} \to 3^3 \to 27\end{align*} ..... 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*}
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Term  Definition 

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
Here you will learn how to calculate the sum of a series when the difference between the members is not constant.