In mathematics, a sequence is an ordered list of numbers. Consider the sequence given by \(1, 4, 9, 16, 25, 36, 49, 64,\cdots\). The ordered part refers to thinking of a sequence as a function, with the domain the positive integers and the value of the sequence term as the range. Sometimes, this function notation is overpowering, so a short hand sequence notation is used:
$$\begin{array}{l} f(1)=1 & \implies & a_1=1 \\ f(2)=4 & \implies & a_2=4 \\ f(3)=9 & \implies & a_3=9 \\ f(4)=16 & \implies & a_4=16 \\ \hfill \vdots & & \vdots \end{array}$$If there are a finite number of terms, say \(n\), then the last term of the sequence would be indicated by \(a_n\). Otherwise, the sequence is an infinite sequence, denoted by the \(\cdots\) notation as used in the above example of squares.
Write out the first 6 terms of the sequence \(a_n=\large{\frac{n+1}{n}}\).
| \(\begin{array}{l} a_1=\large{\frac{1+1}{1}} & \implies & a_1=2 \\ a_2=\large{\frac{2+1}{2}} & \implies & a_2=\large{\frac{3}{2}} \\ a_3=\large{\frac{3+1}{3}} & \implies & a_3=\large{\frac{4}{3}} \\ a_4=\large{\frac{4+1}{4}} & \implies & a_4=\large{\frac{5}{4}} \\ a_5=\large{\frac{5+1}{5}} & \implies & a_5=\large{\frac{6}{5}} \\ a_6=\large{\frac{6+1}{6}} & \implies & a_6=\large{\frac{7}{6}} \end{array}\) |
Find an explicit formula, \(a_n\), for the sequence \(3, 7, 11, 15, 19,\cdots\).
| Try to find a pattern: | \(\underset{+4}{\underset{\smile}{3, 7}} \implies \underset{+4}{\underset{\smile}{7, 11}} \implies \underset{+4}{\underset{\smile}{11, 15}} \implies \underset{+4}{\underset{\smile}{15, 19}}\cdots\) |
| Terms written with pattern: | \(\begin{array}{l} a_1=3 & \implies a_1=3+0 & \implies a_1=3+4(0) \\ a_2=7 & \implies a_2=3+4 & \implies a_2=3+4(1) \\ a_3=11 & \implies a_3=7+4 & \implies a_3=3+4(1)+4 & \implies a_3=3+4(2) \\ a_4=15 & \implies a_4=11+4 & \implies a_4=3+4(2)+4 & \implies a_4=3+4(3) \\ a_5=19 & \implies a_5=15+4 & \implies a_5=3+4(3)+4 & \implies a_3=3+4(4) \end{array}\) |
| The multiplier is one less than the term number: | \(\begin{array}{l} a_\color{blue}{1}=3+4(\color{blue}{0}) \\ a_\color{blue}{2}=3+4(\color{blue}{1}) \\ a_\color{blue}{3}=3+4(\color{blue}{2}) \\ a_\color{blue}{4}=3+4(\color{blue}{3}) \\ a_\color{blue}{5}=3+4(\color{blue}{4}) \\ \hspace{0.25in}\vdots\hspace{0.75in}\vdots \\ a_\color{blue}{n}=3+4(\color{blue}{n-1}) \end{array}\) |
| Write the explicit, or general term: | \(a_n=3+4(n-1)\;\color{green}{\checkmark}\) |
Many sequences are defined recursively, which means the general term uses previous terms to define a formula. A very famous example is called the Fibonacci Sequence, where the general term is found by adding up the two terms in front of it: \(a_n=a_{n-1}+a_{n-2}\). The video details how to find terms using this sequence.
A factorial is a calculation formed by multiplying all the numbers from a given number back to \(1\): \(n!=n\cdot(n-1)\cdot(n-2)\cdots3\cdot2\cdot1\), and for completeness, \(0!=1\). A quick example is \(5!=5\cdot4\cdot3\cdot2\cdot1=120\). The video will also show how to evaluate expressions using factorials.
The idea of a series is to take terms of a sequence and add them together. Since a sequence can have a finite number of terms, or an infinite number of terms, a series can be finite of infinite. When a series if finite, it is called a partial sum. The notation used for adding up terms is called summation notation and looks like: $$\sum_{k=1}^n a_n=a_1+a_2+a_3+\cdots+a_n$$ In the sequence is infinite, then the infinite series would look like: $$\sum_{k=1}^\infty a_n=a_1+a_2+a_3+\cdots+a_j+\cdots$$
Find the sum: \(\displaystyle{\sum_{k=1}^5}\normalsize{\left(1+k^2\right)}\).
| Plug in each value of \(k\): | \(\sum_{k=1}^5\left(1+k^2\right)=\left(1+(\color{blue}{1})^2\right)+\left(1+(\color{blue}{2})^2\right) +\left(1+(\color{blue}{3})^2\right)+\left(1+(\color{blue}{4})^2\right)+\left(1+(\color{blue}{5})^2\right)\) |
| Calculate each step using order of operations: | \(\begin{array}{l} \implies (1+1)+(1+4)+(1+9)+(1+16)+(1+25) \\ \implies 2+5+10+17+26 \\ \implies 60 \\ \implies \sum_{k=1}^5\left(1+k^2\right)=60\;\color{green}{\checkmark} \end{array}\) |
Find the infinite sum: \(\displaystyle{\sum_{k=1}^\infty}\normalsize{2\left(0.10\right)^k}\).
| Plug in a few values of \(k\): | \(\begin{array}{l} \displaystyle{\sum_{k=1}^\infty}\normalsize{2\left(0.10\right)^k}= 2(0.10)^1+2(0.10)^2+2(0.10)^3+2(0.10)^4+\cdots \\ \implies 2(0.10)+2(0.01)+2(0.001)+2(0.0001)+\cdots \\ \implies 0.2+0.02+0.002+0.0002+\cdots \\ \implies 0.2222\cdots \\ \implies \large{\frac{2}{9}} \\ \implies \displaystyle{\sum_{k=1}^\infty}\normalsize{2\left(0.10\right)^k}=\frac{2}{9}\normalsize{\;\color{green}{\checkmark}} \end{array}\) |