Precalculus by Richard Wright

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

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 10^{th} 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.

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

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

- Plug in the lower limit for the index variable.
- Put a plus.
- Plug in the next number for the index variable.
- Plus.
- Plug in the next number, plus, etc. until you plug in the upper limit.

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

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

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.

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

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

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

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

28

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

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

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

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

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

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

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

3135

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

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

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

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

Start by finding the rule for the *n*^{th} term assuming the lower limit is \(\color{purple}{1}\).

n |
1 | 2 | 3 | 4 | 5 |
---|---|---|---|---|---|

a_{n} |
4 | 5 | 6 | 7 | 8 |

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

Now find the upper limit using the rule for the *n*^{th} 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.

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

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

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

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

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

- Plug in the lower limit for the index variable.
- Put a plus.
- Plug in the next number for the index variable.
- Plus.
- Plug in the next number, plus, etc. until you plug in the upper limit.

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

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

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

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

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

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

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

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

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

Fill in the pieces of the summation: index, lower limit, upper limit, rule for *n*th 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.

- \(\displaystyle \sum_{i=1}^{4} 3i\)
- \(\displaystyle \sum_{i=2}^{6} i^2 + 1\)
- \(\displaystyle \sum_{i=1}^{3} \frac{i}{10}\)
- \(\displaystyle \sum_{i=3}^{5} 2^i\)
- \(\displaystyle \sum_{n=1}^{6} 3(2)^{n-1}\)
- \(\displaystyle \sum_{k=2}^{4} k^3\)
- \(\displaystyle \sum_{i=1}^{∞} \frac{3}{10^i}\)
- \(\displaystyle \sum_{n=1}^{∞} 5\left(\frac{1}{10}\right)^n\)
- \(\displaystyle \sum_{i=1}^{30} i^2\)
- \(\displaystyle \sum_{i=1}^{14} 2i - i^2\)
- \(\displaystyle \sum_{n=1}^{10} n^5 + n^3\)
- \(5 + 8 + 11 + 14 + \cdots + 29\)
- \(1 + 4 + 9 + 16 + \cdots + 144\)
- \(6 + 18 + 54 + 162 + \cdots + 39366\)
- 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 10
^{th}generation? - (10-01) Write the rule for the
*n*^{th}term: 3, 7, 11, 15, … - (10-01) Write the first 4 terms of the sequence: \(a_n = -4\left(\frac{1}{2}\right)^{n-1}\)
- (9-06) Use Cramer's Rule to solve: \(\left\{\begin{align} 2x - 4y = 3 \\ x + 3y = 4 \end{align}\right.\)
- (9-04) Find the inverse of \(\left[\begin{matrix} 1 & 3 & 1 \\ 0 & -2 & 2 \\ 0 & 0 & 1 \end{matrix}\right]\)
- (6-03) Find the magnitude of \(\langle 4, -3 \rangle\).

Evaluate the summation.

Evaluate the infinite summation using partial fractions.

Evaluate the summations using the shortcut formulas.

Write the series as a summation.

Problem Solving

Mixed Review

- 30
- 95
- \(\frac{3}{5}\)
- 56
- 189
- 99
- \(\frac{1}{3}\)
- \(\frac{5}{9}\)
- 9455
- -805
- 223850
- \(\displaystyle \sum_{i=1}^{9} (3i+2)\)
- \(\displaystyle \sum_{i=1}^{12} i^2\)
- \(\displaystyle \sum_{i=1}^{9} 2·3^i\)
- 2046 people
- \(a_n = 4n - 1\)
- \(-4, -2, -1, -\frac{1}{2}\)
- \(\left(\frac{5}{2}, \frac{1}{2}\right)\)
- \(\left[\begin{matrix} 1 & \frac{3}{2} & -4 \\ 0 & -\frac{1}{2} & 1 \\ 0 & 0 & 1 \end{matrix}\right]\)
- 5