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# Sums of Finite Geometric Series

## Series with defined ending value has sum: Sn = (a1(1-r^n))/(1-r)

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Sums of Finite Geometric Series

Anna is on a progressive workout plan, so every day she adds 5% to her exercise time. If she starts by exercising 15 mins 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: 15+(151.05)+[(151.05)1.05]...\begin{align*}15 + (15 \cdot 1.05) + [(15 \cdot 1.05) \cdot 1.05]...\end{align*} and so on up to 45, but that would be horribly tedious. In this lesson, you will learn how to answer a question like this will little effort.

### Sums of Finite Geometric Series

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 S2 = 75

The sum of the first three is 50 + 25 + 12.5 = 87.5. We can write this as S3 = 87.5

To find the value of Sn 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 nth term is an = a1rn - 1 , we can use induction to prove a formula for Sn .

The sum of the first n terms in a geometric series is Sn=a1(1rn)1r\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:

Sn=a1(1rn)1r=50(1(12)6)112=50(1164)12=50(6364)12=50(6364)(21)=98716\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 TI-83/4 calculator:

We can use this formula as long as the series in question is geometric.

### Examples

#### Example 1

Earlier, you were asked a question about Anna and her progressive workout plan.

Every day she adds 5% to her exercise time. If she starts by exercising 15 mins on the first day, how many minutes will she have exercised all together on day 45?

Use the formula: Sn=a1(1rn)1r\begin{align*}S_{n}=\frac{a_1(1 - r^n)} {1 - r}\end{align*}

Sn=15(11.0545)11.05\begin{align*}S_{n} = \frac{15 (1 - 1.05^{45})}{1 - 1.05}\end{align*}

Sn=2395.5\begin{align*}S_{n} = 2395.5\end{align*} minutes.

#### Example 2

Find the sum of the first 10 terms of a geometric series with a1 = 3 and r = 5.

The sum is 58,593.

Sn=a1(1rn)1r=3(157)15=3(178,125)4=3(78,124)4=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 3

Find the sum of each series:

1. The first term of a geometric series is 4, and the common ratio is 3. Find S8.

S8=4(138)13=13,120\begin{align*}S_{8}=\frac {4(1-3^8)}{1-3}=13,120\end{align*}

1. The first term of a geometric series is 80, and the common ratio is (1/4). Find S7.

S7=80(1(14)7)114106.66\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 4

Prove the formula Sn=a1(1rn)1r\begin{align*}S_{n}=\frac{a_1(1 - r^n)} {1 - r}\end{align*} by induction.

Step 1) If n = 1, the nth sum is the first sum, or a1 . Using the hypothesized equation, we get S1=a1(1r1)1r=a1(1r)1r=a1\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.

Step 2) Assume that the sum of the first k terms in a geometric series is Sk=a1(1rk)1r\begin{align*}S_k = \frac{a_1(1 - r^k)} {1 - r}\end{align*}.

Step 3) Show that the sum of the first k+1 terms in a geometric series is Sk+1=a1(1rk+1)1r\begin{align*}S_{k + 1} = \frac{a_1(1 - r^{k + 1})} {1 - r}\end{align*}.

Sk+1=Sk+ak+1\begin{align*}S_{k + 1} = S_k + a_{k + 1}\end{align*} The k+1\begin{align*}k + 1\end{align*} sum is the kth\begin{align*}k^{th}\end{align*} sum, plus the k+1\begin{align*}k + 1\end{align*} term
=a1(1rk)1r+a1rk+11\begin{align*} = \frac{a_1(1 - r^k)} {1 - r} + a_1r^{k + 1 - 1}\end{align*} Substitute from step 2, and substitute the k+1\begin{align*}k + 1\end{align*} term
=a1(1rk)1r+a1rk(1r)1r\begin{align*}= \frac{a_1(1 - r^k)} {1 - r} + \frac{a_1r^k(1 - r)} {1- r}\end{align*} The common denominator is 1r\begin{align*}1 - r\end{align*}
=a1(1rk)+a1rk(1r)1r\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.

#### Example 5

Find the sum: \begin{align*}5 + 10 + 20 +...+ 640\end{align*} (Hint: if an = 640 , what is n?).

\begin{align*}S_{8}=\frac {5(1-2^8)}{1-2} = 1275\end{align*}

#### Example 6

Write the first 5 terms of the sequence: \begin{align*}-5 \cdot \frac{3}{4}^n\end{align*}.

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*}

#### Example 7

Write the 3rd, 4th, and 6th terms of: \begin{align*}(3)^{\left(\frac{n}{2}\right)}\end{align*}.

As with Example 6, 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*}

#### Example 8

Find the sum of the series: \begin{align*}\sum_{n = 1}^6 \left(-\frac{3}{2}\right)^{n-1}\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*}

### Review

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*}.

1. \begin{align*}1 + \left(-\frac{1}{2}\right) + \frac{1}{4} + ... + \frac{1}{64}\end{align*}
2. \begin{align*}-6 + 12 -24 + ... -6144\end{align*}
3. \begin{align*}(-4) + (-12) + (-36) + ... + (-2916)\end{align*}
4. \begin{align*}9 + (-45) + 225 + ... + (-17,578,125)\end{align*}
5. \begin{align*}\sum_{n=1}^5 -(2)^{n-1}\end{align*}
6. \begin{align*}\sum_{n=1}^{10} 6\left(\frac{1}{2}\right)^{n-1}\end{align*}
7. \begin{align*}\sum_{n=1}^6 8 \cdot 3^{n-1}\end{align*}
8. \begin{align*}(-5) + 10 + (-20) + ... + (-1280)\end{align*}
9. \begin{align*}(-9) + \frac{9}{2} + \left(-\frac{9}{4}\right) + ... + \left(-\frac{9}{16}\right)\end{align*}
10. \begin{align*}\sum_{n=1}^9 4 \cdot \left(\frac{1}{2}\right)^{n-1}\end{align*}
11. \begin{align*}\sum_{n=1}^{10} 4 \cdot \left(-\frac{2}{3}\right)^{n-1}\end{align*}
12. \begin{align*}(-3) + \left(-\frac{3}{2}\right) + \left(-\frac{3}{4}\right) + ... + \left(-\frac{3}{1024}\right)\end{align*}
13. \begin{align*}\sum_{n=1}^{11} -9 \cdot (-2)^{(n-1)}\end{align*}
14. \begin{align*}\sum_{n=1}^9 -9 \cdot \left(-\frac{5}{3}\right)^{n-1}\end{align*}
15. \begin{align*}\sum_{n=1}^6 -2 \cdot \left(-\frac{5}{4}\right)^{n-1}\end{align*}
16. \begin{align*}\sum_{n=1}^{11} 8 \cdot \left(-\frac{1}{3}\right)^{n-1}\end{align*}

To see the Review answers, open this PDF file and look for section 7.9.

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### Vocabulary Language: English

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.