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:
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
Note that every term is made up of a 1, and a set of 2’s. How many 2’s are in each term?
a_{1} | = 1 |
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a_{2} | = 1 + 2 = 3 |
a_{3} | = 1 + 2 × 2 = 5 |
a_{4} | = 1 + 3 × 2 = 7 |
a_{5} | = 1 + 4 × 2 = 9 |
a_{6} | = 1 + 5 × 2 = 11 |
The n^{th} term has (n - 1) 2's. For example, a_{99} = 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) |
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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} | = a_{1} + d(n - 1) |
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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?
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.
a_{n} = 4n - 1 , and a_{50} = 199
The first term of the sequence is 3, and the common difference is 4.
a_{n} | = a_{1} + d(n - 1) |
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a_{n} | = 3 + 4(n - 1) |
a_{n} | = 3 + 4n - 4 |
a_{n} | = 4n - 1 |
a_{50} | = 4(50) - 1 = 200 - 1 = 199 |
We can also find an explicit formula for a geometric sequence. Consider the following sequence:
t_{2} = 2t_{1} = 2 × 3 = 6 | ||
---|---|---|
t_{1} = 3 | \begin{align*}\rightarrow\end{align*} |
t_{3} = 2t_{2} = 2 × 6 = 12 |
t_{n} = 2 × t_{n}_{-1} | t_{4} = 2t_{3} = 2 × 12 = 24 | |
t_{5} = 2t_{4} = 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.
t_{1} | = 3 |
---|---|
t_{2} | = 2 × 3 = 6 |
t_{3} | = 2 × 2 × 6 = 2^{2} × 6 = 12 |
t_{4} | = 2 × 2 × 2 × 6 = 2^{3} × 6 = 24 |
t_{5} | = 2 × 2 × 2 × 2 × 6 = 2^{4} × 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(2^{n} ^{- 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} | = t_{1}(r^{n}^{-1}) |
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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.
a_{n} = 5 × 3^{n} ^{- 1}, and a_{10} = 98,415
The first term in the sequence is 5, and r = 3.
a_{n} | = a^{1} × r^{n} ^{- 1} |
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a_{n} | = 5 × 3^{n - 1} |
a_{10} | = 5 × 3^{10 - 1} |
a_{10} | = 5 × 3^{9} = 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)...
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.
For the sequence: 2, 9, 16...
\begin{align*}a_n = 7n - 5\end{align*}
\begin{align*}\therefore a_{20} = 7(20) - 5\end{align*}
\begin{align*}a_{20} = 135\end{align*}
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.
For the sequence: (1/2), (1/4), (1/8)...
\begin{align*}a_n = \frac{1}{2^n}\end{align*}
\begin{align*}\therefore a_7 = \frac{1}{2^7}\end{align*}
\begin{align*}a_7 = \frac{1}{128}\end{align*}
Example 7
Identify all sequences in the previous two examples that are geometric. What is the common ratio in each sequence?
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.
- \begin{align*}-21, -6, 18, -3, 20, -2\end{align*}
−21,−6,18,−3,20,−2 - \begin{align*}0,\frac{-1}{5}, \frac{-2}{5}, \frac{-3}{5}, \frac{-4}{5}, -1\end{align*}
0,−15,−25,−35,−45,−1 - \begin{align*}1, 3, 9, 27, 81, 243\end{align*}
1,3,9,27,81,243 - \begin{align*}2, 9, -2, 1, 18, 2\end{align*}
2,9,−2,1,18,2
Write the first 5 terms of the arithmetic sequence (explicit).
- \begin{align*}a_n = -8 -9(n-1)\end{align*}
an=−8−9(n−1) - \begin{align*}a_n = 6-\frac{2}{3}(n-1)\end{align*}
an=6−23(n−1) - \begin{align*}a_n = 8 + \frac{1}{3}(n-1)\end{align*}
an=8+13(n−1)
Solve the following:
- What are the first five terms of the sequence? \begin{align*}a_n = a_{n-1} - \frac{10}{3}; a_1 = -6\end{align*}
an=an−1−103;a1=−6 - Given the sequence, write a recursive function to generate it: \begin{align*}2, -4, -10, -16, -22, -28\end{align*}
2,−4,−10,−16,−22,−28 - Write the equation of \begin{align*}a_n\end{align*}
an without using recursion: \begin{align*}a_n = a_{n-1} - \frac{3}{2}; a_1 = 10\end{align*}an=an−1−32;a1=10 - Write as a recursion: \begin{align*}a_n = 6 - \frac{5}{3}(n-1)\end{align*}
an=6−53(n−1) - Write the equation of \begin{align*}a_n\end{align*}
an without using recursion: \begin{align*}a_n = a_{n-1}+8; a_1 = 3\end{align*}an=an−1+8;a1=3 - What are the first five terms of the sequence? \begin{align*}a_n = a_{n-1} - 1; a_1 = -5\end{align*}
an=an−1−1;a1=−5
Write an explicit formula for the \begin{align*}n_{th}\end{align*}
- \begin{align*}-7, \frac{-13}{3}, \frac{-5}{3}, 1, \frac{11}{3}, \frac{19}{3}\end{align*}
−7,−133,−53,1,113,193 - \begin{align*}6, -4, -14, -24, -34, -44\end{align*}
6,−4,−14,−24,−34,−44 - \begin{align*}9, 16, 23, 30, 37, 44\end{align*}
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.
- \begin{align*}a_n = 5(-3)^{(n-1)}\end{align*}
an=5(−3)(n−1) - \begin{align*}a_n = -6(\frac{-10}{3}^{(n-1)})\end{align*}
an=−6(−103(n−1))
Write the explicit formula for the \begin{align*}n_{th}\end{align*}
- \begin{align*}-8, 16, -32, 64, -128, 256\end{align*}
−8,16,−32,64,−128,256 - \begin{align*}9, \frac{27}{2}, \frac{81}{4}, \frac{243}{8}, \frac{729}{16}, \frac{2187}{32}\end{align*}
9,272,814,2438,72916,218732
Convert the explicit formula to a recursive formula.
- \begin{align*}a_n = 9(\frac{-4}{3})^{(n-1)}\end{align*}
an=9(−43)(n−1) - \begin{align*}a_n = -6(-4)^{(n-1)}\end{align*}
an=−6(−4)(n−1) - \begin{align*}a_n = -5(5)^{(n-1)}\end{align*}
an=−5(5)(n−1)
Review (Answers)
To see the Review answers, open this PDF file and look for section 7.2.