Why Indexing Matters
A sequence is an ordered list of terms. Its index tells you where each term sits in that order. To read a sequence correctly, you need to know both the rule for its terms and the index at which the sequence begins.
For example, the rule \(a_n=2n-1\) produces different initial terms depending on whether \(n\) starts at \(1\) or at \(0\). The algebraic expression alone does not specify the starting index. Sequence notation supplies that information, and careful attention to it prevents errors when listing terms or comparing formulas.
Reading Sequence Notation
The notation \(a_n\) names the term at index \(n\). The notation \(\{a_n\}_{n=1}^{\infty}\) indicates the sequence of terms \(a_1,a_2,a_3,\ldots\). More generally, \(\{a_n\}_{n=m}^{\infty}\) begins with \(a_m\), then continues with \(a_{m+1},a_{m+2},\ldots\).
The index is not a term: if the sequence begins at \(n=1\), its first term is \(a_1\), not \(a_0\). In a formula, substitute the actual index value into every occurrence of \(n\). Then simplify to find the term.
Some sequences are defined starting at \(n=0\). In that case, \(a_0\) is the first term. Do not assume every sequence starts at \(1\); check the lower index or the stated domain before calculating.
Worked Example: Listing Terms from an Explicit Formula
Let \(a_n=n^2+1\) for \(n\geq 1\). Find the first four terms.
The first four indices are \(1,2,3,4\). Substitute each index into the formula:
Therefore the first four terms are \(2,5,10,17\). Checking the substitutions against the original rule confirms that each term is its index squared plus \(1\).
Explicit Formulas and Index Shifts
An explicit formula gives a term directly from its index. To find \(a_{20}\), for example, you can substitute \(20\) into the formula without first finding the earlier terms. A formula may still depend on the index range: its starting index determines which term is first.
An index shift changes which index is assigned to the first term. Compare the same expression \(2n-1\) under two different starting indices:
The expression is identical, but the first index differs. Thus the initial term differs too. If the index begins at \(0\), substituting \(n=0\) gives \(2(0)-1=-1\); if it begins at \(1\), substituting \(n=1\) gives \(2(1)-1=1\).
Index shifts can also be expressed by changing the formula. For instance, the sequence \(1,3,5,7,\ldots\) can be indexed from \(1\) by \(a_n=2n-1\), or from \(0\) by \(b_n=2n+1\). In the second description, \(b_0=1\), \(b_1=3\), and \(b_2=5\). The sequence of values is the same; the labels on those values have changed.
Worked Example: Comparing Two Indexing Conventions
Suppose \(a_n=3n+2\) for \(n\geq 1\), while \(b_n=3n+5\) for \(n\geq 0\). List the first three terms of each sequence and compare them.
For \(a_n\), the first three indices are \(1,2,3\):
For \(b_n\), the first three indices are \(0,1,2\):
Both lists begin \(5,8,11\). The check is direct: corresponding terms have the same values, even though their indices differ by \(1\). In fact, \(b_n=a_{n+1}\) for \(n\geq 0\), since \(a_{n+1}=3(n+1)+2=3n+5=b_n\).
Recursive Rules
A recursive rule defines one or more initial terms and then gives a rule for producing later terms from earlier ones. Unlike an explicit formula, a recurrence may not give a term directly from its index. To calculate a later term, you often need to calculate the terms before it.
For example, \(a_1=4\) together with \(a_n=a_{n-1}+3\) for \(n\geq 2\) defines each term after the first by adding \(3\) to the previous term. The condition \(n\geq 2\) matters: it tells us where the recurrence starts and ensures that the referenced term \(a_{n-1}\) has already been defined.
Worked Example: Generating Terms from a Recurrence
Let \(c_1=2\) and \(c_n=2c_{n-1}+1\) for \(n\geq 2\). Find \(c_1,c_2,c_3,c_4\).
The initial value is given, so \(c_1=2\). Apply the recurrence in order:
Thus the first four terms are \(2,5,11,23\). Check the recurrence at each step: \(2(2)+1=5\), \(2(5)+1=11\), and \(2(11)+1=23\), as required.
A recurrence must provide enough initial terms for its rule to work. For a rule involving \(a_{n-2}\), for example, you need starting values for both \(a_1\) and \(a_2\) if the recurrence begins at \(n=3\). A recurrence without the necessary initial value(s) can describe many possible sequences rather than one determined sequence.
From a Recurrence to an Explicit Formula
Some simple recurrences can be rewritten as explicit formulas. This connection explains how repeated use of a recurrence builds a pattern in the index. Consider \(a_1=4\) and \(a_n=a_{n-1}+3\) for \(n\geq 2\). Repeated substitution gives:
At index \(n\), the starting value \(4\) has been increased by \(3\) a total of \(n-1\) times. This suggests the explicit formula \(a_n=4+3(n-1)\). The following derivation verifies it for every allowed index.
At \(n=1\), the proposed expression gives \(4+3(1-1)=4\), which agrees with \(a_1=4\).
Suppose \(a_k=4+3(k-1)\) for some \(k\geq 1\).
Then \(a_{k+1}=a_k+3=[4+3(k-1)]+3=4+3k=4+3((k+1)-1)\). This is the proposed formula at index \(k+1\).
The formula holds at the initial index and passes from any index \(k\) to the next index \(k+1\). By mathematical induction, it holds for every integer \(n\geq 1\). As a check, at \(n=4\) it gives \(a_4=4+3(4-1)=13\); the recurrence gives \(4,7,10,13\), agreeing with that value.
A Common Indexing Mistake
A frequent error is to use the correct formula with the wrong first index. Suppose \(a_n=2n+1\) for \(n\geq 0\), and you are asked for the first three terms. A wrong calculation might begin at \(n=1\): \(a_1=3\), \(a_2=5\), \(a_3=7\), giving \(3,5,7\). That list omits the actual first term because the sequence begins at \(n=0\).
The correction is to read the index range first, then substitute its values in order: \(a_0=2(0)+1=1\), \(a_1=2(1)+1=3\), and \(a_2=2(2)+1=5\). The first three terms are \(1,3,5\). A useful habit is to write the first few indices explicitly before evaluating the formula.
Choosing the Right Reading Strategy
When interpreting sequence notation, separate three pieces of information: the index range, the rule, and the requested term or list. For an explicit formula, substitute the relevant index. For a recursive rule, begin with the stated initial values and apply the recurrence in order. Keep the index attached to each calculated value so that you do not confuse a term with its position.
| Feature | Explicit formula | Recursive rule |
|---|---|---|
| How a term is found | Substitute its index directly | Use initial value(s) and earlier term(s) |
| Information needed | Formula and index range | Initial value(s), recurrence, and index range |
| Typical notation | \(a_n=3n+2\) | \(a_1=5,\ a_n=a_{n-1}+3\) |
| Common risk | Starting at the wrong index | Skipping or miscalculating an earlier term |
These distinctions matter throughout the study of sequences and series. A misplaced starting index can change the initial terms, while a missing initial value can leave a recurrence incomplete. Read the notation literally, identify the starting index, and show the intermediate terms whenever the rule depends on earlier terms.
Check Your Understanding
Use the index ranges and rules stated in each question.
- For \(u_n=4n-1\) with \(n\geq 1\), find the first four terms.
- For \(v_n=n^2\) with \(n\geq 0\), find the first three terms. Which term is \(v_0\)?
- Given \(w_1=3\) and \(w_n=w_{n-1}+5\) for \(n\geq 2\), calculate \(w_1,w_2,w_3,w_4\).
- Explain why the formula \(x_n=x_{n-1}+2\) alone does not specify one sequence.
- Rewrite the recurrence \(p_1=6,\ p_n=p_{n-1}+4\) for \(n\geq 2\) as an explicit formula, and check it at \(n=3\).
