AP Calculus BC glossary

Recursive Sequence

Also called: Recursive formula, Recurrence relation

A recursive sequence is given by one or more starting values together with a rule that produces each new term from the terms before it, instead of an explicit formula in n. Unless you can find a closed form, and arithmetic and geometric recursions always have one, you reach a distant term by iterating.

a_{n+1} = f(a_n, a_{n-1}, \ldots), \qquad a_1, \ldots, a_k \ \text{given}

Recursive definitions are not an AP Calculus topic in their own right. Reading a recurrence off a list of opening terms belongs to AP Precalculus; on this exam recursion arrives through Euler's method, Topic 7.5, where each new value is built from the previous one. In Unit 10 the term-to-term view is the useful one, since a geometric series has $a_{n+1} = r\,a_n$ and the ratio test looks straight at $a_{n+1}/a_n$ .

How many starting values you need is set by how far back the rule reaches. A first order rule takes one: $a_{n+1} = 2a_n + 1$ with $a_1 = 2$ gives $2, 5, 11, 23, \ldots$ . Fibonacci reaches back two terms, $a_{n+1} = a_n + a_{n-1}$ , so it needs two seeds, $a_1 = a_2 = 1$ .

Iterating is mechanical, and the counting is where marks go missing. Reaching $a_5$ from $a_1$ takes four applications of the rule, not five. Euler's method has the identical shape: each new value comes from the previous value plus the step size times the slope there.

The mistake

Substituting $n$ into a recursive rule as though the rule itself were closed. For $a_{n+1} = a_n + 3$ with $a_1 = 2$ a closed form does exist, $a_n = 3n - 1$ , and the tenth term is $2 + 3(9) = 29$ , since nine steps separate the first term from the tenth. Writing $2 + 3(10)$ counts one step too many.

Appears in: Unit 10: Infinite Sequences and Series (BC)