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

Mr. Wright teaches the lesson.

Summary: In this section, you will:

SDA NAD Content Standards (2018): PC.7.2

Bible
Bible study. (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 + ⋯

Summation Notation

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.

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

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

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

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.

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.

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

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.

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

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

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

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

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

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

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.

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

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

Write a Summation from a Series

Write the summation notation for 4 + 5 + 6 + 7 + 8 + ⋯ + 15.

Solution

Start by finding the rule for the nth term assuming the lower limit is 1.

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

Notice an = 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 = 12

Now fill in the parts of the summation notation.

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

Write the summation notation for 3 + 5 + 7 + 9 + 11 + ⋯ + 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

  1. Evaluate the summation.
  2. \(\displaystyle \sum_{i=1}^{4} 3i\)
  3. \(\displaystyle \sum_{i=2}^{6} i^2 + 1\)
  4. \(\displaystyle \sum_{i=1}^{3} \frac{i}{10}\)
  5. \(\displaystyle \sum_{i=3}^{5} 2^i\)
  6. \(\displaystyle \sum_{n=1}^{6} 3(2)^{n-1}\)
  7. \(\displaystyle \sum_{k=2}^{4} k^3\)
  8. Evaluate the infinite summation using partial fractions.
  9. \(\displaystyle \sum_{i=1}^{∞} \frac{3}{10^i}\)
  10. \(\displaystyle \sum_{n=1}^{∞} 5\left(\frac{1}{10}\right)^n\)
  11. Evaluate the summations using the shortcut formulas.
  12. \(\displaystyle \sum_{i=1}^{30} i^2\)
  13. \(\displaystyle \sum_{i=1}^{14} 2i - i^2\)
  14. \(\displaystyle \sum_{n=1}^{10} n^5 + n^3\)
  15. Write the series as a summation.
  16. 5 + 8 + 11 + 14 + ⋯ + 29
  17. 1 + 4 + 9 + 16 + ⋯ + 144
  18. 6 + 18 + 54 + 162 + ⋯ + 39366
  19. Problem Solving
  20. 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?
  21. Mixed Review
  22. (10-01) Write the rule for the nth term: 3, 7, 11, 15, …
  23. (10-01) Write the first 4 terms of the sequence: \(a_n = -4\left(\frac{1}{2}\right)^{n-1}\)
  24. (9-06) Use Cramer's Rule to solve: \(\left\{\begin{align} 2x - 4y = 3 \\ x + 3y = 4 \end{align}\right.\)
  25. (9-04) Find the inverse of \(\left[\begin{matrix} 1 & 3 & 1 \\ 0 & -2 & 2 \\ 0 & 0 & 1 \end{matrix}\right]\)
  26. (6-03) Find the magnitude of ⟨4, −3⟩.

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. an = 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