Precalculus by Richard Wright

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Husbands, love your wives, just as Christ loved the church and gave himself up for her. Ephesians‬ ‭5‬:‭25‬ ‭NIV‬‬‬

# 10-02 Series

Summary: In this section, you will:

• Evaluate a summation.
• Write a series as a summation.

SDA NAD Content Standards (2018): PC.7.2

Let's say you are a religious person. At some point when you are young, you win 2 people for God. Each of those win 2 more people for God. Assuming that each person continues to win 2 more people, thus doubling the number of people. How many people will you be indirectly responsible for winning for God after the 10th generation?

This follows a sequence where the terms double, 2, 4, 8, 16, 32, 64, …; however, this asks for the total number of people. That means the terms need to be added together: 2 + 4 + 8 + 16 + 32 + 64 + …. The sum of a sequence is a series.

$$Sequence: 2, 4, 8, 16, …$$
$$Series: 2 + 4 + 8 + 16 + \cdots$$

A special way of writing series called summation notation, or sigma notation, is used to shorten the way a series is written. It uses the Greek capital letter sigma, Σ.

###### Summation Notation (Sigma Notation)

$$\sum_{\text{index}=\text{lower limit}}^{\text{upper limit}} (\text{sequence explicit rule})$$

$$\sum_{i=1}^{n} a_i$$

###### Evaluate a Summation
1. Plug in the lower limit for the index variable.
2. Put a plus.
3. Plug in the next number for the index variable.
4. Plus.
5. Plug in the next number, plus, etc. until you plug in the upper limit.

$$\sum_{i=1}^{n} a_i = a_1 + a_2 + a_3 + \cdots + a_n$$

#### Example 1: Evaluate a Summation

Find the sum $$\displaystyle \sum_{i=1}^{4} (2i - 1)$$.

###### Solution

The lower limit is 1, so start by plugging in a 1, then +, then plug in 2, +, continue until you plug in the upper limit 4.

$$\sum_{i=1}^{4} (2i - 1) = (2\color{blue}{(1)} - 1) \color{red}{+} (2\color{blue}{(2)} - 1) \color{red}{+} (2\color{blue}{(3)} - 1) \color{red}{+} (2\color{blue}{(4)} - 1)$$

$$= 1 + 3 + 5 + 7 = 16$$

#### Example 2: Evaluate a Summation

Find the sum $$\displaystyle \sum_{k=3}^{5} (k^2 + k)$$.

###### Solution

The lower limit is 3, so start by plugging in a 3, then +, then plug in 4, +, continue until you plug in the upper limit 5.

$$\sum_{k=3}^{5} \left(k^2 + k\right) = \left(\left(\color{blue}{3}\right)^2 + \left(\color{blue}{3}\right)\right) \color{red}{+} \left(\left(\color{blue}{4}\right)^2 + \left(\color{blue}{4}\right)\right) \color{red}{+} \left(\left(\color{blue}{5}\right)^2 + \left(\color{blue}{5}\right)\right)$$

$$= 12 + 20 + 30 = 62$$

##### Try It 1

Find the sum $$\displaystyle \sum_{i=2}^{5} (2i)$$

28

###### Infinite Series

Some infinite series can be calculated. Do this by finding a pattern in the partial sums.

1. Evaluate the first term.
2. Add the first 2 terms.
3. Add the first 3 terms.
4. Add the first 4 terms.
5. Continue until a pattern is seen.

#### Example 3: Evaluate an Infinite Series

Evaluate $$\displaystyle \sum_{n=1}^{∞} \frac{2}{10^n}$$.

###### Solution

Since this is an infinite sum, find the partial sums until a pattern is discovered. The first term is

$$\frac{2}{10^{\color{blue}{1}}} = 0.2$$

Find the 2nd partial sum by adding the first 2 terms.

$$\frac{2}{10^{\color{blue}{1}}} + \frac{2}{10^{\color{blue}{2}}} = 0.2 + 0.02 = 0.22$$

Find the 3rd partial sum by adding the first 3 terms.

$$\frac{2}{10^{\color{blue}{1}}} + \frac{2}{10^{\color{blue}{2}}} + \frac{2}{10^{\color{blue}{3}}} = 0.2 + 0.02 + 0.002 = 0.222$$

Find the 4th partial sum by adding the first 4 terms.

$$\frac{2}{10^{\color{blue}{1}}} + \frac{2}{10^{\color{blue}{2}}} + \frac{2}{10^{\color{blue}{3}}} + \frac{2}{10^{\color{blue}{4}}} = 0.2 + 0.02 + 0.002 = 0.0002 = 0.2222$$

The pattern appears to be repeating 2's.

$$\sum_{n=1}^{∞} \frac{2}{10^n} = 0.\bar{2} = \frac{2}{9}$$

##### Try It 2

Evaluate $$\displaystyle \sum_{n=1}^{∞} 7\left(\frac{1}{10}\right)^n$$.

$$\frac{7}{9}$$

###### Shortcut Summation Formulas

Use these formulas when the lower limit is 1. Simply plug in the upper limit for n.

$$\sum_{i=1}^{n} i = 1 + 2 + 3 + 4 + 5 + \cdots + n = \frac{n(n+1)}{2}$$

$$\sum_{i=1}^{n} i^2 = 1^2 + 2^2 + 3^2 + 4^2 + 5^2 + \cdots + n^2 = \frac{n(n+1)(2n+1)}{6}$$

$$\sum_{i=1}^{n} i^3 = 1^3 + 2^3 + 3^3 + 4^3 + 5^3 + \cdots + n^3 = \frac{n^2(n+1)^2}{4}$$

$$\sum_{i=1}^{n} i^4 = 1^4 + 2^4 + 3^4 + 4^4 + 5^4 + \cdots + n^4 = \frac{n(n+1)(2n+1)(3n^2+3n-1)}{30}$$

$$\sum_{i=1}^{n} i^5 = 1^5 + 2^5 + 3^5 + 4^5 + 5^5 + \cdots + n^5 = \frac{n^2(n+1)^2(2n^2+2n-1)}{12}$$

#### Example 4: Evaluate a Summation using a Formula

Evaluate $$\displaystyle \sum_{i=1}^{20} (2i^2 - i)$$.

###### Solution

Because addition has the associative property, the summation can be split into two summations.

$$\sum_{i=1}^{20} (2i^2 - i) = \sum_{i=1}^{20} 2\color{blue}{i^2} - \sum_{i=1}^{20} \color{red}{i}$$

Now fill in the formulas.

$$2\left(\color{blue}{\frac{n(n+1)(2n+1)}{6}}\right) - \left(\color{red}{\frac{n(n+1)}{2}}\right)$$

Because the upper limit is 20, substitute n = 20 and simplify.

$$2\left(\frac{\color{purple}{20}(\color{purple}{20}+1)(2(\color{purple}{20})+1)}{6}\right) - \left(\frac{\color{purple}{20}(\color{purple}{20}+1)}{2}\right)$$

$$= 5530$$

##### Try It 3

Evaluate $$\displaystyle \sum_{i=1}^{10} (i^3 + 2i)$$.

3135

###### Write a Summation from a Series

Fill in the pieces of the summation: index, lower limit, upper limit, rule for nth term.

$$\sum_{\text{index}=\text{lower limit}}^{\text{upper limit}} (\text{sequence explicit rule})$$

$$\sum_{i=1}^{n} a_i$$

#### Example 5: Write a Summation from a Series

Write the summation notation for $$4 + 5 + 6 + 7 + 8 + \cdots + 15$$.

###### Solution

Start by finding the rule for the nth term assuming the lower limit is $$\color{purple}{1}$$.

 n an 1 2 3 4 5 4 5 6 7 8

Notice $$a_n = \color{blue}{n + 3}$$.

Now find the upper limit using the rule for the nth term. Set the rule equal to the final term and solve for n.

$$n + 3 = 15$$

$$n = \color{red}{12}$$

Now fill in the parts of the summation notation.

$$\sum_{n=\color{purple}{1}}^{\color{red}{12}} \color{blue}{n + 3}$$

##### Try It 4

Write the summation notation for $$3 + 5 + 7 + 9 + 11 + \cdots + 21$$.

$$\displaystyle \sum_{n=1}^{10} (2n + 1)$$

##### Lesson Summary

###### Summation Notation (Sigma Notation)

$$\sum_{\text{index}=\text{lower limit}}^{\text{upper limit}} (\text{sequence explicit rule})$$

$$\sum_{i=1}^{n} a_i$$

###### Evaluate a Summation
1. Plug in the lower limit for the index variable.
2. Put a plus.
3. Plug in the next number for the index variable.
4. Plus.
5. Plug in the next number, plus, etc. until you plug in the upper limit.

$$\sum_{i=1}^{n} a_i = a_1 + a_2 + a_3 + \cdots + a_n$$

###### Infinite Series

Some infinite series can be calculated. Do this by finding a pattern in the partial sums.

1. Evaluate the first term.
2. Add the first 2 terms.
3. Add the first 3 terms.
4. Add the first 4 terms.
5. Continue until a pattern is seen.

###### Shortcut Summation Formulas

Use these formulas when the lower limit is 1. Simply plug in the upper limit for n.

$$\sum_{i=1}^{n} i = 1 + 2 + 3 + 4 + 5 + \cdots + n = \frac{n(n+1)}{2}$$

$$\sum_{i=1}^{n} i^2 = 1^2 + 2^2 + 3^2 + 4^2 + 5^2 + \cdots + n^2 = \frac{n(n+1)(2n+1)}{6}$$

$$\sum_{i=1}^{n} i^3 = 1^3 + 2^3 + 3^3 + 4^3 + 5^3 + \cdots + n^3 = \frac{n^2(n+1)^2}{4}$$

$$\sum_{i=1}^{n} i^4 = 1^4 + 2^4 + 3^4 + 4^4 + 5^4 + \cdots + n^4 = \frac{n(n+1)(2n+1)(3n^2+3n-1)}{30}$$

$$\sum_{i=1}^{n} i^5 = 1^5 + 2^5 + 3^5 + 4^5 + 5^5 + \cdots + n^5 = \frac{n^2(n+1)^2(2n^2+2n-1)}{12}$$

###### Write a Summation from a Series

Fill in the pieces of the summation: index, lower limit, upper limit, rule for nth term.

$$\sum_{\text{index}=\text{lower limit}}^{\text{upper limit}} (\text{sequence explicit rule})$$

$$\sum_{i=1}^{n} a_i$$

## Practice Exercises

Evaluate the summation.

1. $$\displaystyle \sum_{i=1}^{4} 3i$$
2. $$\displaystyle \sum_{i=2}^{6} i^2 + 1$$
3. $$\displaystyle \sum_{i=1}^{3} \frac{i}{10}$$
4. $$\displaystyle \sum_{i=3}^{5} 2^i$$
5. $$\displaystyle \sum_{n=1}^{6} 3(2)^{n-1}$$
6. $$\displaystyle \sum_{k=2}^{4} k^3$$
7. Evaluate the infinite summation using partial fractions.

8. $$\displaystyle \sum_{i=1}^{∞} \frac{3}{10^i}$$
9. $$\displaystyle \sum_{n=1}^{∞} 5\left(\frac{1}{10}\right)^n$$
10. Evaluate the summations using the shortcut formulas.

11. $$\displaystyle \sum_{i=1}^{30} i^2$$
12. $$\displaystyle \sum_{i=1}^{14} 2i - i^2$$
13. $$\displaystyle \sum_{n=1}^{10} n^5 + n^3$$
14. Write the series as a summation.

15. $$5 + 8 + 11 + 14 + \cdots + 29$$
16. $$1 + 4 + 9 + 16 + \cdots + 144$$
17. $$6 + 18 + 54 + 162 + \cdots + 39366$$
18. Problem Solving

19. Let's say you are a religious person. At some point when you are young, you win 2 people for God. Each of those win 2 more people for God. Assuming that each person continues to win 2 more people, thus doubling the number of people. How many people will you be indirectly responsible for winning for God after the 10th generation?
20. Mixed Review

21. (10-01) Write the rule for the nth term: 3, 7, 11, 15, …
22. (10-01) Write the first 4 terms of the sequence: $$a_n = -4\left(\frac{1}{2}\right)^{n-1}$$
23. (9-06) Use Cramer's Rule to solve: \left\{\begin{align} 2x - 4y = 3 \\ x + 3y = 4 \end{align}\right.
24. (9-04) Find the inverse of $$\left[\begin{matrix} 1 & 3 & 1 \\ 0 & -2 & 2 \\ 0 & 0 & 1 \end{matrix}\right]$$
25. (6-03) Find the magnitude of $$\langle 4, -3 \rangle$$.

1. 30
2. 95
3. $$\frac{3}{5}$$
4. 56
5. 189
6. 99
7. $$\frac{1}{3}$$
8. $$\frac{5}{9}$$
9. 9455
10. -805
11. 223850
12. $$\displaystyle \sum_{i=1}^{9} (3i+2)$$
13. $$\displaystyle \sum_{i=1}^{12} i^2$$
14. $$\displaystyle \sum_{i=1}^{9} 2·3^i$$
15. 2046 people
16. $$a_n = 4n - 1$$
17. $$-4, -2, -1, -\frac{1}{2}$$
18. $$\left(\frac{5}{2}, \frac{1}{2}\right)$$
19. $$\left[\begin{matrix} 1 & \frac{3}{2} & -4 \\ 0 & -\frac{1}{2} & 1 \\ 0 & 0 & 1 \end{matrix}\right]$$
20. 5