A sequence is a list of numbers that have a pattern defining how each number is found. A quick illustration is the sequnce of perfect squares: \(1,\;4,\;9,\;16,\;25,\cdots\). The \(1\) represents the first term, \(4\) the second term, \(9\) the third term, and so on. A notation to represent that is defined as follows:
| First Term | Second Term | Third Term | Forth Term | General Term | |
|---|---|---|---|---|---|
| Notation | \(a_1\) | \(a_2\) | \(a_3\) | \(a_4\) | \(a_n\) |
| Example (Squares) | \(a_1=1\) | \(a_2=4\) | \(a_3=9\) | \(a_4=16\) | \(a_n=n^2\) |
The first of two special sequences to study is called an arithmetic sequence. This sequence of numbers is special because the same number is added to the previous term to get the next term. This number is called the common difference, and is denoted by \(d\). The terms of the sequence thus look like:
$$a_1,\;a_2=a_1+d,\;a_3=a_2+d,\;a_4=a_3+d,\;\cdots a_n=a_{n-1}+d$$This is called a recursive formula, meaning to find a term, the previous term is needed. For example, to find the tenth term, the ninth term is needed, but then to get the ninth term, the eight term is needed, the seventh term, etc., to actually find the value of \(a_{10}\). Luckily, a formula to find any term of an arithmetic sequence is not that challenging to find:
$$\begin{array}{c|c|c|cc} a_1 & a_2 & a_3 & a_4 & \cdots \\ a_1 & a_2=a_1+d & a_3=\color{blue}{a_2}+d & a_4=\color{blue}{a_3}+d & \cdots \\ a_1 & a_2=a_1+d & a_3=\color{blue}{a_1+d}+d & a_4=\color{blue}{a_1+d+d}+d & \cdots \\ a_1 & a_2=a_1+\color{green}{1}\cdot d & a_3=a_1+\color{green}{2}\cdot d & a_4=a_1+\color{green}{3}\cdot d & \cdots \end{array}$$Each term above shows a calculation that starts with \(a_1\) and then adds \(d\) one less time than the term number. For example, \(a_2\) adds \(d\;1\) time, \(a_3\) adds \(d\;2\) times, \(a_4\) adds \(d\;3\) times, each time one less than the term number. Therefore, the general formula for this arithmetic sequence is \(a_n=a_1+(n-1)\cdot d\). This formula is called the explicit formula for an arithmetic sequence.
Find an explicit formula for the arithmetic sequence: \(-5,\;-2,\;1,\;4,\;7,\;10,\cdots\). Use the formula to find \(a_{25}\).
| First find \(a_1\): | The first term is \(a_1=-5\). |
| Next, find \(d\): | \(\begin{array}{c|c|c|c|c} -5+d=-2 & -2+d=1 & 1+d=4 & 4+d=7 & 7+d=10 \\ d=3 & d=3 & d=3 & d=3 & d=3 \end{array}\) |
| Explicit formula: | \(a_n=-5+(n-1)\cdot 3\;\color{green}{\checkmark}\) |
| Find \(a_{25}\) plug in \(n=25\): | \(\begin{array}{l} a_{25}=-5+(25-1)\cdot3 \\ \implies a_{25}=-5+(24)\cdot3 \\ \implies a_{25}=-5+72 \\ \implies a_{25}=67\;\color{green}{\checkmark} \end{array}\) |
The second special sequence to study is called a geometric sequence. This sequence of numbers is special because the same number is multiplied to the previous term to get the next term. This number is called the common ratio, and is denoted by \(r\). The terms of the sequence thus look like:
$$a_1,\;a_2=a_1\cdot r,\;a_3=a_2\cdot r,\;a_4=a_3\cdot r,\;\cdots a_n=a_{n-1}\cdot r$$To find an explicit formula for a geometric sequence, note the pattern of repeated multiplication:
$$\begin{array}{l|l|l|l} a_1 & & & a_\color{green}{1}\cdot r^{\color{green}{0}} \\ a_2 & a_2=\color{blue}{a_1\cdot r} & & a_\color{green}{2}=a_1\cdot r^{\color{green}{1}} \\ a_3 & a_3=a_2\cdot r & a_3=\color{blue}{a_1\cdot r}\cdot r & a_\color{green}{3}=a_1\cdot r^{\color{green}{2}} \\ a_4 & a_4=a_3\cdot r & a_4=\color{blue}{a_1\cdot r^2}\cdot r & a_\color{green}{4}=a_1\cdot r^{\color{green}{3}} \end{array}$$Comparing the term number with the exponent on the common ratio: \(a_1\) multiplies by \(r\;0\) times, \(a_2\;1\) time, \(a_3\;2\) times, \(a_4\;3\) times, so the pattern emerges that the exponent is one less than the term number: \(a_{\color{red}{k}}=a_1\cdot r^{\color{red}{k-1}}\). This leads to the explicit formula:
$$a_n=a_1\cdot r^{n-1}$$Find an explicit formula for the geometric sequence: \(1,\;6,\;36,\;216,\;1296,\cdots\). Use the explicit formula to find the \(32^{nd}\) term.
The explicit formula needs two parts: \(a_1\) and \(r\). From the sequence, \(a_1=1\). To find \(r\), set up the equations that give the multiplication and solve for \(r\) by division:
$$\begin{array}{c} 1\cdot r=6 & \implies r=6 \\ 6\cdot r=36 & \implies r=6 \\ 36\cdot r=216 & \implies r=6 \\ 216\cdot r=1296 & \implies r=6 \end{array}$$The first term of the sequence is given, \(a_1=1\). Therefore, the explicit formula for this geometric sequence is given by: \(a_n=1\left(6\right)^{n-1}\).
To find the \(32^{nd}\) term, plug in \(n=32\) for the explicit formula:
$$a_{32}=1\left(6\right)^{32-1}\implies a_{32}=1\left(6\right)^{31}=1.326443518E24$$Recall this is scientific notation and represents the number \(1.326443518\times 10^{24}\).