Two descriptions, one sequence
A recursive formula defines a sequence in two parts: a starting value (or values) and a rule that builds each new term from the term(s) before it. An explicit formula, by contrast, gives aₙ directly as a function of the position n, with no need to know earlier terms.
Same sequence, two ways: 3, 7, 11, 15, 19, ... Recursive: a_1 = 3, a_n = a_(n-1) + 4 (each term = previous + 4) Explicit: a_n = 3 + (n - 1)(4) = 4n - 1 Recursive must walk: a_4 = a_3 + 4 = 11 + 4 = 15 Explicit jumps: a_50 = 4(50) - 1 = 199 (no need to find a_49 first)
When each one shines
Recursive formulas are natural when each step is defined in terms of the last — interest added to a balance, a population multiplying, the famous Fibonacci rule aₙ = aₙ₋₁ + aₙ₋₂. They are easy to state but slow to query: to get the 100th term you must compute all 99 before it.
Explicit formulas are the opposite: harder to discover, but they let you substitute any n and land on that term instantly. For arithmetic and geometric sequences we already have clean explicit forms, so we usually prefer them for direct calculation.
Converting recursive to explicit
- Read the recursive rule. If it says aₙ = aₙ₋₁ + d, the sequence is arithmetic with common difference d. If it says aₙ = aₙ₋₁ · r, it is geometric with common ratio r.
- Read off the first term a₁ from the starting value.
- Plug a₁ and d (or r) into the matching explicit formula: aₙ = a₁ + (n − 1)d for arithmetic, or aₙ = a₁ · r^(n − 1) for geometric, then simplify.
Convert: a_1 = 6, a_n = (1/2) * a_(n-1) This is geometric: r = 1/2, a_1 = 6 Explicit: a_n = a_1 * r^(n - 1) = 6 * (1/2)^(n - 1) Check: a_3 = 6 * (1/2)^2 = 6 * 1/4 = 3/2 List: 6, 3, 3/2, 3/4, ... (each is half the last — correct)