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# Explicit Formulas

## Identify and apply a general rule for terms in a sequence

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Using and Writing nth Term Rules for Sequences

You buy new furniture at zero percent interest on a monthly installment plan. The total of your furniture is \$4800. The following sequence shows the balance you still owe on the furniture at the beginning of each month. How would you write a general rule for the sequence?

4800, 4600, 4400, 4200,...

### Guidance

In the previous concept we wrote a recursive rule to find the next term in a sequence. Recursive rules can help us generate multiple sequential terms in a sequence but are not helpful in determining a particular single term. Consider the sequence: 3,5,7,,an\begin{align*} 3, 5, 7, \ldots, a_n \end{align*}. The recursive rule for this sequence is an=an1+2\begin{align*}a_n=a_{n-1}+2\end{align*}. What if we want to find the 100th\begin{align*}100^{th}\end{align*} term? The recursive rule only allows us to find a term in the sequence if we know the previous term. An nth\begin{align*}n^{th}\end{align*} term or general rule, however, will allow us to find the 100th\begin{align*}100^{th}\end{align*} term by replacing n\begin{align*}n\end{align*} in the formula with 100.

#### Example A

Write the first three terms, the 15th\begin{align*}15^{th}\end{align*} term and the 40th\begin{align*}40^{th}\end{align*} term of the sequence with the general rule: an=n21\begin{align*}a_n=n^2-1\end{align*}.

Solution: We can find each of these terms by replacing n\begin{align*}n\end{align*} with the appropriate term number:

a1a2a3a15a40=(1)21=0=(2)21=3=(3)21=8=(15)21=224=(40)21=1599

Calculator: These terms can also be found using a graphing calculator. First press 2nd\begin{align*}2^{nd}\end{align*} STAT (to get to the List menu). Arrow over to OPS, select option 5: seq( and type in (expression, variable, begin, end). For this particular problem, the calculator yields the following:

seq(x21,x,1,3)={0 3 8}\begin{align*}seq\left(x^2-1,x,1,3 \right) =\{0 \ 3 \ 8\}\end{align*} for the first three terms

seq(x21,x,15,15)={224}\begin{align*}seq\left(x^2-1,x,15,15 \right)=\{224\}\end{align*} for the 15th\begin{align*}15^{th}\end{align*} term

seq(x21,x,40,40)={1599}\begin{align*}seq\left(x^2-1,x,40,40 \right)=\{1599\}\end{align*} for the 40th\begin{align*}40^{th}\end{align*} term

#### Example B

Write a general rule for the sequence: 5,10,15,20,\begin{align*} 5, 10, 15, 20,\ldots\end{align*}

Solution: The previous example illustrates how a general rule maps a term number directly to the term value. Another way to say this is that the general rule expresses the nth\begin{align*}n^{th}\end{align*} term as a function of n\begin{align*}n\end{align*}. Let’s put the terms in the above sequence in a table with their term numbers to help identify the rule.

Looking at the terms and term numbers together helps us to see that each term is the result of multiplying the term number by 5. The general rule is an=5n\begin{align*}a_n=5n\end{align*}

n\begin{align*}n\end{align*} 1 2 3 4
a\begin{align*}a\end{align*} 5 10 15 20

#### Example C

Find the nth\begin{align*}n^{th}\end{align*} term rule for the sequence: 0,2,6,12,\begin{align*}0, 2, 6, 12,\ldots\end{align*}

Solution: Let’s make the table again to begin to analyze the relationship between the term number and the term value.

n\begin{align*}n\end{align*} 1 2 3 4
an\begin{align*}a_n\end{align*} 0 2 6 12
n(?)\begin{align*}n(?)\end{align*} (1)(0)\begin{align*}(1)(0)\end{align*} (2)(1)\begin{align*}(2)(1)\end{align*} (3)(2)\begin{align*}(3)(2)\end{align*} (4)(3)\begin{align*}(4)(3)\end{align*}

This time the pattern is not so obvious. To start, write each term as a product of the term number and a second factor. Then it can be observed that the second factor is always one less that the term number and the general rule can be written as an=n(n1)\begin{align*}a_n=n(n-1)\end{align*}

Intro Problem Revisit Let’s put the terms in the sequence in a table with their term numbers to help identify the rule.

n\begin{align*}n\end{align*} 1 2 3 4
a\begin{align*}a\end{align*} 4800 4600 4400 4200

Looking at the terms and term numbers together helps us to see that each term is the result of subtracting 200 times one less than the term from the first term. The general rule is an=an1200(n1)\begin{align*}a_n=a_{n-1} - 200(n - 1)\end{align*}.

### Guided Practice

1. Given the general rule: an=3n13\begin{align*}a_n=3n-13\end{align*}, write the first five terms, 25th\begin{align*}25^{th}\end{align*} term and the 200th\begin{align*}200^{th}\end{align*} term of the sequence.

2. Write the general rule for the sequence: 4,5,6,7,\begin{align*}4, 5, 6, 7,\ldots\end{align*}

3. Write the general rule and find the 35th\begin{align*}35^{th}\end{align*} term of the sequence: 1,0,3,8,15,\begin{align*}-1, 0, 3, 8, 15,\ldots\end{align*}

1. Plug in the term numbers as shown:

a1a2a3a4a5a25a200=3(1)13=10=3(2)13=7=3(3)13=4=3(4)13=1=3(5)13=2=3(25)13=62=3(200)13=587

2. Put the values in a table with the term numbers and see if there is a way to write the term as a function of the term number.

n\begin{align*}n\end{align*} 1 2 3 4
an\begin{align*}a_n\end{align*} 4 5 6 7
n±(?)\begin{align*}n\pm(?)\end{align*} (1)+3\begin{align*}(1)+3\end{align*} (2)+3\begin{align*}(2)+3\end{align*} (3)+3\begin{align*}(3)+3\end{align*} (4)+3\begin{align*}(4)+3\end{align*}

Each term appears to be the result of adding three to the term number. Thus, the general rule is \begin{align*}a_n=n+3\end{align*}

3. Put the values in a table with the term numbers and see if there is a way to write the term as a function of the term number.

\begin{align*}n\end{align*} 1 2 3 4 5
\begin{align*}a_n\end{align*} -1 10 3 8 15
\begin{align*}n(?)\end{align*} \begin{align*}(1)(-1)\end{align*} \begin{align*}(2)(0)\end{align*} \begin{align*}(3)(1)\end{align*} \begin{align*}(4)(2)\end{align*} \begin{align*}(5)(3)\end{align*}

Each term appears to be the result of multiplying the term number by two less than the term number. Thus, the general rule is \begin{align*}a_n=n(n-2)\end{align*}.

### Vocabulary

\begin{align*}n^{th}\end{align*} term or general rule
A formula which relates the term to the term number and thus can be used to calculate any term in a sequence whether or not any terms are known.

### Explore More

Use the \begin{align*}n^{th}\end{align*} term rule to generate the indicated terms in each sequence.

1. \begin{align*}2n+7\end{align*}, terms \begin{align*}1-5\end{align*} and the \begin{align*}10^{th}\end{align*} term.
2. \begin{align*}-5n-1\end{align*}, terms \begin{align*}1-3\end{align*} and the \begin{align*}50^{th}\end{align*} term.
3. \begin{align*}2^n-1\end{align*}, terms \begin{align*}1-3\end{align*} and the \begin{align*}10^{th}\end{align*} term.
4. \begin{align*}\left(\frac{1}{2}\right)^n\end{align*}, terms \begin{align*}1-3\end{align*} and the \begin{align*}8^{th}\end{align*} term.
5. \begin{align*}\frac{n(n+1)}{2}\end{align*}, terms \begin{align*}1-4\end{align*} and the \begin{align*}20^{th}\end{align*} term.

Use your calculator to generate the first 5 terms in each sequence. Use MATH > FRAC, on your calculator to convert decimals to fractions.

1. \begin{align*}4n-3\end{align*}
2. \begin{align*} -\frac{1}{2}n+5\end{align*}
3. \begin{align*}\left(\frac{2}{3}\right)^n+1\end{align*}
4. \begin{align*}2n(n-1)\end{align*}
5. \begin{align*}\frac{n(n+1)(2n+1)}{6}\end{align*}

Write the \begin{align*}n^{th}\end{align*} term rule for the following sequences.

1. \begin{align*}3,5,7,9,\ldots\end{align*}
2. \begin{align*}1,7,25,79,\ldots\end{align*}
3. \begin{align*}6,14,24,36,\ldots\end{align*}
4. \begin{align*}6,5,4,3,\ldots\end{align*}
5. \begin{align*}2,5,9,14,\ldots\end{align*}

### Vocabulary Language: English

arithmetic sequence

arithmetic sequence

An arithmetic sequence has a common difference between each two consecutive terms. Arithmetic sequences are also known are arithmetic progressions.
common difference

common difference

Every arithmetic sequence has a common or constant difference between consecutive terms. For example: In the sequence 5, 8, 11, 14..., the common difference is "3".
common ratio

common ratio

Every geometric sequence has a common ratio, or a constant ratio between consecutive terms. For example in the sequence 2, 6, 18, 54..., the common ratio is 3.
geometric sequence

geometric sequence

A geometric sequence is a sequence with a constant ratio between successive terms. Geometric sequences are also known as geometric progressions.
index

index

The index of a term in a sequence is the term’s “place” in the sequence.