Chapter 8 - Sequences
| Thursday, 2 September 2021 | |
| 5-minute read | |
| 877 words | |
A sequence is a collection of terms that follows a general rule, or an equation, which develops a pattern.
- term - element of a sequence
General Rule
\[ a_n = \frac{(-1)^n}{2n - 1} \]
Where \(n\) is the term number, and \(n\) must be a positive integer.
\[ a_1 = \frac{(-1)^1}{2(1) - 1} = -1 \],
\[ a_2 = \frac{(-1)^2}{2(2) - 1} = \frac{1}{3} \],
\[ a_3 = \frac{(-1)^3}{2(3) - 1} = -\frac{1}{5} \],
\[ a_4 = \frac{(-1)^4}{2(4) - 1} = \frac{1}{7} \],
\[ a_5 = \frac{(-1)^5}{2(5) - 1} = -\frac{1}{9} \],
Factorial List
\[ 0! = 1 \]
\[ 1! = 0!(1) = 1 \]
\[ 2! = 1!(2) = 2 \]
\[ 3! = 2!(3) = 6 \]
\[ 4! = 3!(4) = 24 \]
\[ 5! = 4!(5) = 120 \]
\[ 6! = 5!(6) = 720 \]
Examples
- Write an expression for the nth term of each sequence
\[ 1, 3, 5, 7, ... = 2n - 1 \]
\[ -2, 1, 6, 13, 22 = x^2 - 3 \]
\[ \frac{1}{2}, \frac{1}{2}, \frac{3}{4}, \frac{3}{2}, \frac{15}{4}, \frac{45}{4} = \frac{n!}{2^n} \]
- Simplify these expressions
\[ \frac{10!}{7!} = 10 \cdot 9 \cdot 8 = 720 \]
\[ \frac{(n - 1)!}{(n + 1)!} = \frac{1}{n(n + 1)} = \frac{1}{n^2 + n} \]
Fibonacci Sequence
\[ a_n = a_{n - 1} + a_{n - 2}, a_1 = 1, a_2 = 2 \]
Arithmetic Sequences
Arithmetic sequences have a common difference between each term that adds or subtracts to the next term.
General explicit rule: \(a_n = a_1 + d(n - 1)\)
General recursive rule: \(a_n = a_{n - 1} + d\)
Examples
- Find the general rule \(a_n\) for the sequence 5, 8, 11, 14, …
\[ d = 3 \]
\[ a_n = d(n - 1) + 5 \]
\[ a_n = 3n + 2 \]
- Find the general rule \(a_n\) for the sequence \(a_1, a_1 + d, a_1 + 2d\)
\[ a_n = d(n - 1) + a_1 \]
- What is the 4th term of an arithmetic sequence whose first term is 7 and common difference is -2?
\[ a_n = -2(n - 1) + 7 \]
\[ a_4 = -2(4 - 1) + 7 = 1 \]
- What is the 872nd term of the previous arithmetic sequence?
\[ a_{872} = -2(871) + 7 = -1742 + 7 = -1735 \]
Partial Sum of an Arithmetic Sequence
\[ S_n = \frac{n(a_1 + a_n)}{2} \]
Geometric Sequences
Geometric sequences have a common ratio between each term that multiplies to the next term.
If \(|r| > 1\) then the sequence will generally increase. If \(0 < |r| < 1\), then the sequence will generally decrease. If \(r < 0\) then the sequence will alternate signs.
General explicit rule: \(a_n = a_1 \cdot r^{n - 1}\)
General recursive rule: \(a_n = r \cdot a_{n - 1}\)
Finding Common Ratio Given Two Terms
\[ r = \sqrt[m - n]{\frac{a_m}{a_n}} \]
Examples
- What is the general rule for the geometric sequence 2, 14, 98, …
\[ a_n = 2 \cdot 7^{n - 1} \]
- Find a_10 for the geometric sequence where \(r = \frac{1}{2}\) and \(a_4 = 125\)
\[ a_{10} = a_4 \cdot \frac{1}{2}^6 = \frac{125}{64} \]
Partial Sum of a Geometric Sequence
\[ S_n = a_1(\frac{1 - r^n}{1 - r}) \]
Example
- Evaluate \(\sum_{n = 1}^{12} 4(0.3)^n\)
\[ S_n = 1.2(\frac{1 - 0.3^{12}}{1 - 0.3}) = 1.714 \]
Sum of an Infinite Geometric Series
Given \(|r| < 1\)
\[ S_{\infty} = \frac{a_1}{1 - r} \]
Proof
Given the geometric sequence \(a_n = a_1 \cdot r^{n - 1}\)
Examples
- Evaluate \(\sum_{n = 1}^{\infty} 4(\frac{1}{2})^n\)
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\(\frac{2}{\frac{1}{2}} = 4\), therefore this series converges to 4.
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- Evaluate \(\sum_{n = 1}^{\infty} 4(2)^n\)
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\(|r| \ge 1\), therefore this series diverges to \(\infty\).
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Proof by Mathematical Induction
Formal Types of Logical Reasoning
Deductive reasoning is a logical process in which a conclusion ins based on the concordance of multiple premises that are generally assumed to be true
Inductive reasoning starts with a generalization, and then test it by applying it to specific incidents.
Process of Inductive Reasoning
- Show that your statement is true for its initial occurrence.
- Make the assumption that it will be true for any arbitrarily selected trial of the rule.
- Prove that it will be true for the term after the arbitrarily selected trial.
Example
- Prove that \(1 + 3 + 5 + 7 + ... + (2k - 1) = k^2\)
Prove for 1st term \(k = 1\)
\[ 2(1) - 1 = 1^2 \]
Assume \(1 + 3 + 5 + 7 + ... + (2k - 1) = k^2\)
Prove for \(k + 1\)
\[ 1 + 3 + 5 + 7 + ... + (2k - 1) + (2(k + 1) - 1) = (k + 1)^2 \]
- Prove that \(1 + 4 + 9 + 16 + ... + n^2 = \frac{n(n + 1)(2n + 1)}{6}\)
Prove for 1st term \(n = 1\)
\[ 1^2 = \frac{(2)(3)}{6} \]
Assume \(1 + 4 + 9 + 16 + ... + n^2 = \frac{n(n + 1)(2n + 1)}{6}\)
Prove for \(n + 1\)
$$ 1 + 3 + 5 + 7 + … + n^2 + (n + 1)^2 = \frac{(n + 1)(n + 2)(2n + 3)}{6}