We have evaluated explicit formulas for sequences, including arithmetic and geometric sequences. Explicit formulas, if you recall, allow us to find any term in the sequence if we know its position, or \(n\) value.
A recursive rule does not operate the same way. In order to find the \(nth\) term in the sequence, you need to know the term or terms that precede it. We depict these prior terms using notation such as \(a\mathstrut{_{n-1}}\) or \(a_{n-2}\), etc. The subscript, \(n - 1\), means one term prior in the sequence. The subscript \(n - 2\) means two terms prior in the sequence. We can also depict terms using \(a_{n+1}\) or \(a_{n+2}\), etc. In this case the subscript \(n+1\) means one term AFTER term \(n\).
For example, if \(n = 5\), \(a_{n-1}\) is equivalent to \(a_{5-1}=a_{4}\). Just as with explicit rules, a value is substituted for the variable \(n\) when implementing the formula.
In order to list terms of a recursively defined sequence, it is necessary to be given the first term (or terms) depending on how the formula is written.
Example 1: List the first 5 terms of the recursively defined sequence given by the formula.
\(\begin{cases} a_1=5\\ a_n=3a_{n-1}+n\end{cases} \)
The first term is given: \(5\)
The second term is found by substituting \(2\) into the formula for every \(n\) value: \(a_2=3a_{2-1}+2\).
Simplifying we get \(a_2=3a_1+2\). Since \(a_1=5\) we have \(a_2=3(5)+2=15+2=17\)
The third term is found as follows:
\(a_3=3a_{3-2}+3=3a_2+3=3(17)+3=51+3=54\)
Once you establish the pattern, you can quickly find the remaining terms:
\(a_4=3a_3+4=3(54)+4=162+4=166\)
\(a_5=3a_4+5=3(166)+5=498+5=503\)
Thus, the first five terms of the recursively defined sequence are \(5,\ 17,\ 54,\ 166,\ 503\).
Example 2: List the first \(5\) terms of the recursively defined sequence given by \(\begin{cases} a_1=-2\\ a_2=3 \\a_n=(-1)^n \cdot a_{n-1}-a_{n-2} \end{cases}\)
A recursive rule does not operate the same way. In order to find the \(nth\) term in the sequence, you need to know the term or terms that precede it. We depict these prior terms using notation such as \(a\mathstrut{_{n-1}}\) or \(a_{n-2}\), etc. The subscript, \(n - 1\), means one term prior in the sequence. The subscript \(n - 2\) means two terms prior in the sequence. We can also depict terms using \(a_{n+1}\) or \(a_{n+2}\), etc. In this case the subscript \(n+1\) means one term AFTER term \(n\).
For example, if \(n = 5\), \(a_{n-1}\) is equivalent to \(a_{5-1}=a_{4}\). Just as with explicit rules, a value is substituted for the variable \(n\) when implementing the formula.
In order to list terms of a recursively defined sequence, it is necessary to be given the first term (or terms) depending on how the formula is written.
Example 1: List the first 5 terms of the recursively defined sequence given by the formula.
\(\begin{cases} a_1=5\\ a_n=3a_{n-1}+n\end{cases} \)
The first term is given: \(5\)
The second term is found by substituting \(2\) into the formula for every \(n\) value: \(a_2=3a_{2-1}+2\).
Simplifying we get \(a_2=3a_1+2\). Since \(a_1=5\) we have \(a_2=3(5)+2=15+2=17\)
The third term is found as follows:
\(a_3=3a_{3-2}+3=3a_2+3=3(17)+3=51+3=54\)
Once you establish the pattern, you can quickly find the remaining terms:
\(a_4=3a_3+4=3(54)+4=162+4=166\)
\(a_5=3a_4+5=3(166)+5=498+5=503\)
Thus, the first five terms of the recursively defined sequence are \(5,\ 17,\ 54,\ 166,\ 503\).
Example 2: List the first \(5\) terms of the recursively defined sequence given by \(\begin{cases} a_1=-2\\ a_2=3 \\a_n=(-1)^n \cdot a_{n-1}-a_{n-2} \end{cases}\)
Example 3: List the first \(4\) terms of the recursively defined sequence given by \(\begin{cases} a_1=-2\\ a_{n+1}=3a_n-7n \end{cases} \).
The first term is given: \(-2\)
To find the second term, which is \(a_2\), figure out the \(n\) value by solving \(a_2=a_{n+1}\) for \(n\). This means that \(n = 1\).
Therefore, \(a_2=3a_1-7\left(1\right)=3\left(-2\right)-7=-6-7=-13\).
To find the third term, it means that \(n = 2\), so \(a_3=3a_2-7\left(2\right)=3\left(-13\right)-7\left(2\right)=-39-14=-53\).
The fourth term is \(a_4=3a_3-7\left(3\right)=3\left(-53\right)-21=-159-21=-180\).
The first four terms are \(-2,\ -13,\ -53,\ -180\).
Recursive Formula for Arithmetic Sequences
Arithmetic sequences, if you recall, have a common difference, \(d\), between terms. The recursive rule for an arithmetic sequence is \(\begin{cases} a_1=value\\ a_n=a_{n-1}+d\end{cases}\).
Example 4: Write a recursive formula for the arithmetic sequence \(10,\ 7,\ 4,\ 1,\ ...\)
Since the first term is \(10\) and the difference between terms is \(-3\), the formula is \(\begin{cases} a_1=10\\ a_n=a_{n-1}-3\end{cases}\).
Example 5: Using the recursively defined formula \(\begin{cases} a_1=-2\\ a_n=a_{n-1}+2\end{cases}\), write the explicit rule for the arithmetic sequence.
Recursive Formula for Geometric Sequences
Geometric sequences, if you recall, have a common ratio, \(r\), between terms. The recursive rule for a geometric sequence is \(\begin{cases} a_1=value\\ a_n=r \cdot a_{n-1}\end{cases}\).
Example 6: Write a recursive formula for the geometric sequence \(-5,\ 10,\ -20,\ 40,\ ...\)
Since the first term is \(-5\) and the common ratio is \(-2\), the formula is \(\begin{cases} a_1=-5\\ a_n=-2\cdot a_{n-1}\end{cases}\).