<meta http-equiv="refresh" content="1; url=/nojavascript/">
You are viewing an older version of this Concept. Go to the latest version.

# Sums of Finite Geometric Series

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

0%
Progress
Practice Sums of Finite Geometric Series
Progress
0%
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: $15 + (15 \cdot 1.05) + [(15 \cdot 1.05) \cdot 1.05]...$ 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.

Embedded Video:

### 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 $S_n = \frac{a_1(1 - r^n)} {1 - r}$

For example, for the series 50 + 25 + 12.5 + ... , the sum of the first 6 terms is:

$S_n = \frac{a_1(1 - r^n)} {1 - r} = \frac{50 \left(1 - (\frac{1} {2})^6 \right)} {1 - \frac{1} {2}} = \frac{50 \left(1 - \frac{1} {64} \right)} {\frac{1} {2}} = \frac{50 \left(\frac{63} {64} \right)} {\frac{1} {2}} = 50 \left(\frac{63} {64} \right) \left(\frac{2} {1} \right) = 98 \frac{7} {16}$

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.

#### 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.

$S_n = \frac{a_1(1 - r^n)} {1 - r} = \frac{3(1 - 5^7)} {1 - 5} = \frac{3(1 - 78,125)} {-4} = \frac{3(-78,124)} {-4} = 58,593$
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) $S_{8}=\frac {4(1-3^8)}{1-3}=13,120$
b) $S_{7}= \frac {80 \left ( {1- \left ( \frac{1}{4} \right )^7} \right )}{1-\frac {1}{4}} \approx 106.66$

#### Example C

Prove the formula $S_{n}=\frac{a_1(1 - r^n)} {1 - r}$ by induction

Solution

1. If n = 1, the n th sum is the first sum, or a 1 . Using the hypothesized equation, we get $S_1 = \frac{a_1(1 - r^1)} {1 - r} = \frac{a_1(1 - r)} {1 - r} = a_1$ . This establishes the base case.
2. Assume that the sum of the first k terms in a geometric series is $S_k = \frac{a_1(1 - r^k)} {1 - r}$ .
3. Show that the sum of the first k +1 terms in a geometric series is $S_{k + 1} = \frac{a_1(1 - r^{k + 1})} {1 - r}$ .
$S_{k + 1} = S_k + a_{k + 1}$ The $k + 1$ sum is the $k^{th}$ sum, plus the $k + 1$ term
$= \frac{a_1(1 - r^k)} {1 - r} + a_1r^{k + 1 - 1}$ Substitute from step 2, and substitute the $k + 1$ term
$= \frac{a_1(1 - r^k)} {1 - r} + \frac{a_1r^k(1 - r)} {1- r}$ The common denominator is $1 - r$
$= \frac{a_1(1 - r^k) + a_1r^k(1 - r)} {1 - r}$ Simplify the fraction
$= \frac{a_1 \left[1 - r^k + r^k(1 - r) \right]} {1 - r}$
$= \frac{a_1 \left[1 - r^k + r^k - r^{k + 1} \right]} {1 - r}$
$= \frac{a_1 \left[1 - r^{k + 1} \right]} {1 - r}$ It is proven.

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

Concept question wrap-up "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: $S_{n}=\frac{a_1(1 - r^n)} {1 - r}$

$S_{n} = \frac{15 (1 - 1.05^{45})}{1 - 1.05}$
$S_{n} = 2395.5$ minutes.

### 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: $5 + 10 + 20 +...+ 640$ (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: $-5 \cdot \frac{3}{4}^n$

4) Write the 3rd, 4th, and 6th terms of: $(3)^{\left(\frac{N}{2}\right)}$

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

Solutions

1) $S_{8}=\frac {5(1-2^8)}{1-2} = 1275$

2) $S_{12}=\frac {11000(1-1.05^{12})}{1-1.05} = 175,088.39$

3) Just do the multiplication for each term $n = 0 \to n = 4$

$-5 \cdot \frac{3}{4}^0 \ to -5 \cdot 1 \to -5$ ..... for $n = 0$
$-5 \cdot \frac{3}{4}^1 \ to -5 \cdot \frac{3}{4} \to -\frac{15}{4} \to -3.75$ ..... for $n = 1$
$-5 \cdot \frac{3}{4}^2 \ to -5 \cdot \frac{9}{16} \to -\frac{45}{16} \to -2.8$ ..... for $n = 2$
$-5 \cdot \frac{3}{4}^3 \ to -5 \cdot \frac{27}{64} \to -\frac{135}{64} \to -2.1$ ..... for $n = 3$
$-5 \cdot \frac{3}{4}^4 \ to -5 \cdot \frac{81}{256} \to -\frac{405}{256} \to -1.6$ ..... for $n = 4$
$\therefore$ 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 :

$3^{\frac{3}{2}} \to \sqrt{3^3} \to \sqrt{27} \to 5.2$ ..... for $n = 3$
$3^{\frac{4}{2}} \to 9$ ..... for $n = 4$
$3^{\frac{6}{2}} \to 3^3 \to 27$ ..... for $n = 6$
$\therefore$ the 3rd, 4th, and 6th terms are: $5.2, 9, 27$

5) To find the sum of the series $\sum_{n = 1}^6 \left(-\frac{3}{2}\right)^{n-1}$

We could calculate all of the values for $n = 1 \to 6$ and add them, getting:
$1+ \frac{-3}{2} + \frac{9}{4} + \frac{-27}{8} + \frac{81}{16} + \frac{-243}{32} = \frac{-133}{32}$
Or we can use the formula: $\left( \frac{1 - r^k}{1 - r} \right)$
$\left( \frac{1 - \left(\frac{-3}{2}\right)^6}{1 - \left(\frac{-3}{2}\right)} \right) = \frac{-133}{32}$

### Practice

Find the sum of the finite series. You may simply calculate the individual terms and add them, or you may use the formula: $S_n = \frac{a_1 (1- r^n)}{1- r}$

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

### Vocabulary Language: English

finite series

finite series

A series is finite if it has a defined ending value.
geometric series

geometric series

A geometric series is a geometric sequence written as an uncalculated sum of terms.
induction

induction

Induction is a method of mathematical proof typically used to establish that a given statement is true for all positive integers.
infinite series

infinite series

An infinite series is the sum of the terms in a sequence that has an infinite number of terms.
series

series

A series is the sum of the terms of a sequence.