Precalculus by Richard Wright

Previous Lesson Table of Contents Next Lesson

Are you not my student and
has this helped you?

This book is available
to download as an epub.


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

10-02 Series

Mr. Wright teaches the lesson.

Summary: In this section, you will:

SDA NAD Content Standards (2018): PC.7.2

Bible
Figure 1: Bible study. credit (pixabay/Free-Photos)

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 \(\displaystyle \sum_{i=2}^{5} (2i)\)

Answers

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

Answers

\(\frac{7}{9}\)

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

Answers

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 1 2 3 4 5
an 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 \(3 + 5 + 7 + 9 + 11 + \cdots + 21\).

Answers

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

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.

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

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

Helpful videos about this lesson.

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 \(\langle 4, -3 \rangle\).

Answers

  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