7.1.2: Explicit Formulas

Last updated
Save as PDF

Page ID: 14783

\( \newcommand{\vecs}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} } \) \( \newcommand{\vecd}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash {#1}}} \)\(\newcommand{\id}{\mathrm{id}}\) \( \newcommand{\Span}{\mathrm{span}}\) \( \newcommand{\kernel}{\mathrm{null}\,}\) \( \newcommand{\range}{\mathrm{range}\,}\) \( \newcommand{\RealPart}{\mathrm{Re}}\) \( \newcommand{\ImaginaryPart}{\mathrm{Im}}\) \( \newcommand{\Argument}{\mathrm{Arg}}\) \( \newcommand{\norm}[1]{\| #1 \|}\) \( \newcommand{\inner}[2]{\langle #1, #2 \rangle}\) \( \newcommand{\Span}{\mathrm{span}}\) \(\newcommand{\id}{\mathrm{id}}\) \( \newcommand{\Span}{\mathrm{span}}\) \( \newcommand{\kernel}{\mathrm{null}\,}\) \( \newcommand{\range}{\mathrm{range}\,}\) \( \newcommand{\RealPart}{\mathrm{Re}}\) \( \newcommand{\ImaginaryPart}{\mathrm{Im}}\) \( \newcommand{\Argument}{\mathrm{Arg}}\) \( \newcommand{\norm}[1]{\| #1 \|}\) \( \newcommand{\inner}[2]{\langle #1, #2 \rangle}\) \( \newcommand{\Span}{\mathrm{span}}\)\(\newcommand{\AA}{\unicode[.8,0]{x212B}}\)

Explicit Formulas

Rachel and Elaina have started a website where they debate the best color of hair dye. The show is really popular and the visitors to their website are increasing very rapidly. They figure that membership is increasing by about 500 people every three days.

At this rate, how many members will they have on the 48th day? How many days will it be before they reach 25,000 members?

Explicit Formulas

When we represent a sequence with a formula that lets us find any term in the sequence without knowing any other terms, we are representing the sequence explicitly.

Given a recursive definition of an arithmetic or geometric sequence, you can always find an explicit formula, or an equation to represent the n^th term of the sequence. Consider for example the sequence of odd numbers we started with: 1, 3, 5, 7, ...

We can find an explicit formula for the n^th term of the sequence if we analyze a few terms:

\(\ \begin{array}{l}
a_{1}=1 \\
a_{2}=a_{1}+2=1+2=3 \\
a_{3}=a_{2}+2=1+2+2=5 \\
a_{4}=a_{3}+2=1+2+2+2=7 \\
a_{5}=a_{4}+2=1+2+2+2+2=9 \\
a_{6}=a_{5}+2=1+2+2+2+2+2=11
\end{array}\)

Note that every term is made up of a 1, and a set of 2’s. How many 2’s are in each term?

a₁	= 1
a₂	= 1 + 2 = 3
a₃	= 1 + 2 × 2 = 5
a₄	= 1 + 3 × 2 = 7
a₅	= 1 + 4 × 2 = 9
a₆	= 1 + 5 × 2 = 11

The n^th term has (n - 1) 2's. For example, a₉₉ = 1 + 98 × 2 = 197 . We can therefore represent the sequence as a_n = 1 + 2(n - 1). We can simplify this expression:

a_n	= 1 + 2(n - 1)
a_n	= 1 + 2n - 2
a_n	= 2n - 1

In general, we can represent an arithmetic sequence in this way, as long as we know the first term and the common difference, d. Notice that in the previous example, the first term was 1, and the common difference, d, was 2. The n^th term is therefore the first term, plus d(n - 1):

a_n	= a₁ + d(n - 1)

You can use this general equation to find an explicit formula for any term in an arithmetic sequence.

Examples

Example 1

Earlier, you were asked two questions about a website membership.

If memberships are increasing by about 500 people every three days, how many members will they have on the 48th day? How many days will it be before they reach 25,000 members?

Solution

This is actually a fairly simple arithmetic sequence: each day there are 500/3 more members, on average. Use the formula for arithmetic sequences from the Example 2 below.

Example 2

Find an explicit formula for the nth term of the sequence 3, 7, 11, 15... and use the equation to find the 50^th term in the sequence.

Solution

\(\ a_{n}=4 n-1\), and \(\ a_{50}=199\)

The first term of the sequence is 3, and the common difference is 4.

a_n	= a₁ + d(n - 1)
a_n	= 3 + 4(n - 1)
a_n	= 3 + 4n - 4
a_n	= 4n - 1

a₅₀	= 4(50) - 1 = 200 - 1 = 199

We can also find an explicit formula for a geometric sequence. Consider the following sequence:

		t₂ = 2t₁ = 2 × 3 = 6
t₁ = 3	→	t₃ = 2t₂ = 2 × 6 = 12
t_n = 2 × t_n_-1		t₄ = 2t₃ = 2 × 12 = 24
		t₅ = 2t₄ = 2 × 24 = 48

Notice that every term is the first term, multiplied by a power of 2. This is because 2 is the common ratio for the sequence.

t₁	= 3
t₂	= 2 × 3 = 6
t₃	= 2 × 2 × 6 = 2² × 6 = 12
t₄	= 2 × 2 × 2 × 6 = 2³ × 6 = 24
t₅	= 2 × 2 × 2 × 2 × 6 = 2⁴ × 6 = 48

The power of 2 in the n^th term is (n-1). Therefore the n^th term in this sequence can be defined as: t_n = 3(2ⁿ ^{- 1}). In general, we can define the n^th term of a geometric sequence in terms of its first term and its common ratio, r:

t_n	= t₁(rⁿ^-1)

You can use this general equation to find an explicit formula for any term in a geometric sequence.

Example 3

Find an explicit formula for the n^th term of the sequence 5, 15, 45, 135... and use the equation to find the 10^th term in the sequence.

Solution

a_n = 5 × 3ⁿ ^{- 1}, and a₁₀ = 98,415

The first term in the sequence is 5, and r = 3.

a_n	= a¹ × rⁿ ^{- 1}
a_n	= 5 × 3^{n - 1}
a₁₀	= 5 × 3^{10 - 1}
a₁₀	= 5 × 3⁹ = 5 × 19,683 = 98,415

Again, it is always possible to write an explicit formula for terms of an arithmetic or geometric sequence. However, you can also write an explicit formula for other sequences, as long as you can identify a pattern. To do this, you must remember that a sequence is a function, which means there is a relationship between the input and the output. That is, you must identify a pattern between the term and its index, or the term’s “place” in the sequence.

Example 4

Write an explicit formula for the nth term of the sequence 1, (1/2), (1/3), (1/4)...

Solution

a_n = (1/n)

Initially you may see a pattern in the fractions, but you may also wonder about the first term. If you write 1 as (1/1), then it should become clear that the n^th term is (1/n).

Example 5

Write an explicit formula for the sequence: 2, 9, 16... and use the formula to find the value of the 20^th term.

Solution

For the sequence: 2, 9, 16...

\(\ \begin{array}{l}
a_{n}=7 n-5 \\
\therefore a_{20}=7(20)-5 \\
a_{20}=135
\end{array}\)

Example 6

Write an explicit formula for the sequence: (1/2), (1/4), (1/8) and use the formula to find the value of the 7_th term.

Solution

For the sequence: (1/2), (1/4), (1/8)...

\(\ \begin{array}{l}
a_{n}=\frac{1}{2^{n}} \\
\therefore a_{7}=\frac{1}{2^{7}} \\
a_{7}=\frac{1}{128}
\end{array}\)

Example 7

Identify all sequences in the previous two examples that are geometric. What is the common ratio in each sequence?

Solution

The sequence in Example 5 is arithmetic.

The sequence in Example 6 is geometric and has r = 1/2.

Review

Name the sequence as arithmetic, geometric, or neither.

−21, −6, 18, −3, 20, −2
\(\ 0, \frac{-1}{5}, \frac{-2}{5}, \frac{-3}{5}, \frac{-4}{5},-1\)
1, 3, 9, 27, 81, 243
2, 9, −2, 1, 18, 2

Write the first 5 terms of the arithmetic sequence (explicit).

\(\ a_{n}=-8-9(n-1)\)
\(\ a_{n}=6-\frac{2}{3}(n-1)\)
\(\ a_{n}=8+\frac{1}{3}(n-1)\)

Solve the following:

What are the first five terms of the sequence? \(\ a_{n}=a_{n-1}-\frac{10}{3} ; a_{1}=-6\)
Given the sequence, write a recursive function to generate it: 2, −4, −10, −16, −22, −28
Write the equation of \(\ a_{n}\) without using recursion: \(\ a_{n}=a_{n-1}-\frac{3}{2} ; a_{1}=10\)
Write as a recursion: \(\ a_{n}=6-\frac{5}{3}(n-1)\)
Write the equation of \(\ a_{n}\) without using recursion: \(\ a_{n}=a_{n-1}+8 ; a_{1}=3\)
What are the first five terms of the sequence? \(\ a_{n}=a_{n-1}-1 ; a_{1}=-5\)

Write an explicit formula for the n_th term of the arithmetic sequence.

\(\ -7, \frac{-13}{3}, \frac{-5}{3}, 1, \frac{11}{3}, \frac{19}{3}\)
6, −4, −14, −24, −34, −44
9, 16, 23, 30, 37, 44
In a particular arithmetic sequence, the second term is 4 and the fifth term is 13. Write an explicit formula for this sequence.

Write the first 5 terms of the geometric sequence.

\(\ a_{n}=5(-3)^{(n-1)}\)
\(\ a_{n}=-6\left(\frac{-10^{(n-1)}}{3}\right)\)

Write the explicit formula for the n_th term of the geometric sequence.

−8, 16, −32, 64, −128, 256
\(\ 9, \frac{27}{2}, \frac{81}{4}, \frac{243}{8}, \frac{729}{16}, \frac{2187}{32}\)

Convert the explicit formula to a recursive formula.

\(\ a_{n}=9\left(\frac{-4}{3}\right)^{(n-1)}\)
\(\ a_{n}=-6(-4)^{(n-1)}\)
\(\ a_{n}=-5(5)^{(n-1)}\)

Review (Answers)

To see the Review answers, open this PDF file and look for section 7.2.

Vocabulary

Term	Definition
arithmetic sequence	An arithmetic sequence has a common difference between each two consecutive terms. Arithmetic sequences are also known are arithmetic progressions.
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	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.
Explicit	Explicit formulas define each term in a sequence directly, allowing one to calculate any term in the sequence without knowing the value of the previous terms.
Explicit formula	Explicit formulas define each term in a sequence directly, allowing one to calculate any term in the sequence without knowing the value of the previous terms.
geometric sequence	A geometric sequence is a sequence with a constant ratio between successive terms. Geometric sequences are also known as geometric progressions.
index	The index of a term in a sequence is the term’s “place” in the sequence.
Natural Numbers	The natural numbers are the counting numbers and consist of all positive, whole numbers. The natural numbers are the numbers in the list 1, 2, 3... and are often referred to as positive integers.
recursive	The recursive formula for a sequence allows you to find the value of the n^th term in the sequence if you know the value of the (n-1)^th term in the sequence.
recursive formula	The recursive formula for a sequence allows you to find the value of the n^th term in the sequence if you know the value of the (n-1)^th term in the sequence.
sequence	A sequence is an ordered list of numbers or objects.