# Explicit Formulas for Sequences - Methods, Examples TL;DR An explicit formula for a sequence gives the n-th term a_n directly as a function of n — no need to compute the previous terms first. For an arithmetic sequence, the formula is \[ a_n = a_1 + (n - 1)d \]. For geometric, \[ a_n = a_1 \cdot r^{n-1} \]. This article covers the derivation, the four most-tested sequence types, worked examples, the three off-by-one mistakes that cost marks, and the mathematicians who built the formulas. ## The 1843 Closed-Form Trick That Solved Fibonacci For 643 years after Leonardo Fibonacci described his rabbit-population sequence in 1202, computing the 100th Fibonacci number meant computing all 99 before it. Then in 1843, French mathematician Jacques Binet published a single formula that returned \( F_n \) directly from n, using only the golden ratio and a power: \[ F_n = \frac{1}{\sqrt{5}}\left[ \left( \frac{1 + \sqrt{5}}{2} \right)^n - \left( \frac{1 - \sqrt{5}}{2} \right)^n \right] \] The leap from "compute the previous 99 terms" to "plug n in and get the answer" is the entire point of an **explicit formula** — also called a closed-form formula or the general term. An **explicit formula** for a sequence \( a_n \) is a formula that expresses \( a_n \) as a function of n alone: \( a_n = f(n) \). Given any term position n, you can compute \( a_n \) in one step without computing \( a_1, a_2, \ldots, a_{n-1} \) first. ## Explicit vs Recursive — Two Different Formula Types Sequences can be described two ways: - **Recursive formula.** Each term defined in terms of one or more previous terms. Example: \( a_n = a_{n-1} + 4 \), with \( a_1 = 3 \). To compute \( a_{100} \), you compute \( a_2, a_3, \ldots \) first. - **Explicit formula.** Each term defined directly from its position. Example: \( a_n = 4n - 1 \). To compute \( a_{100} \), plug in \( n = 100 \): \( a_{100} = 4(100) - 1 = 399 \). Both formulas describe the same sequence 3, 7, 11, 15, 19,… The recursive form mirrors the construction; the explicit form returns the answer in one step. | Property | Recursive | Explicit | | --- | --- | --- | | Needs previous terms? | Yes | No | | Easy to derive from pattern? | Often easier | Sometimes harder | | Fast for large n? | No (linear time) | Yes (constant time) | | Common in computer science? | Yes (recursion) | Yes (formulas) | ## How To Find The Explicit Formula — By Sequence Type The recipe depends on the sequence type. The four below cover almost every Grade 9–11 problem. ### Arithmetic sequences Same difference \( d \) between consecutive terms. Formula: \( a_n = a_1 + (n - 1)d \). **Derivation.** Start at \( a_1 \). Each step adds \( d \). After \( n - 1 \) steps, you've added \( d \) a total of \( n - 1 \) times: \( a_n = a_1 + (n - 1)d \). ### Geometric sequences Same ratio \( r \) between consecutive terms. Formula: \( a_n = a_1 \cdot r^{n-1} \). **Derivation.** Same logic — start at \( a_1 \), multiply by \( r \) each step, count \( (n-1) \) steps. ### Quadratic sequences The second differences are constant. Formula: \( a_n = An^2 + Bn + C \) — three coefficients, solved by plugging in any three known terms. ### Fibonacci and Binet's formula Recursive: \( F_n = F_{n-1} + F_{n-2} \) with \( F_1 = F_2 = 1 \). Explicit (Binet's): the closed form above using the golden ratio \( \phi = \frac{1 + \sqrt{5}}{2} \). ## Quick — Standard — Stretch: Three Worked Examples ### Quick — find the explicit formula for 3,7,11,15,19,… Arithmetic with \( a_1 = 3 \) and common difference \( d = 4 \). \[ a_n = 3 + (n - 1)(4) = 3 + 4n - 4 = 4n - 1 \] **Final answer:** \( a_n = 4n - 1 \). Check: \( a_1 = 4(1) - 1 = 3 \). ✓ \( a_5 = 4(5) - 1 = 19 \). ✓ ### Standard (Wrong-Path-First) — find the explicit formula for the geometric sequence 5,15,45,135,… **Wrong path.** First instinct — recognise geometric with ratio 3, then write \( a_n = 5 \cdot 3^n \). Check: \( a_1 = 5 \cdot 3^1 = 15 \). But the first term is 5, not 15. Off by one. **Correct method.** The formula is \( a_n = a_1 \cdot r^{n-1} \), with the exponent \( (n-1) \). \[ a_n = 5 \cdot 3^{n - 1} \] Check: \( a_1 = 5 \cdot 3^0 = 5 \). ✓ \( a_2 = 5 \cdot 3^1 = 15 \). ✓ \( a_4 = 5 \cdot 3^3 = 135 \). ✓ **Final answer:** \( a_n = 5 \cdot 3^{n-1} \). ### Stretch — find the explicit formula for the quadratic sequence 1,4,9,16,25,… The terms are perfect squares — but let's pretend we don't notice and derive the formula from the differences. First differences: 3,5,7,9,… — not constant. Second differences: 2,2,2,… — constant. So the formula has the form \( a_n = An^2 + Bn + C \). Plug in three knowns: \( a_1 = 1 \), \( a_2 = 4 \), \( a_3 = 9 \). - \( a + b + c = 1 \) - \( 4a + 2b + c = 4 \) - \( 9a + 3b + c = 9 \) **Final answer:** \( a_n = n^2 \). The terms are indeed the perfect squares. ## Why Explicit Formulas Matter — From Population Biology To Algorithm Runtime Explicit formulas show up wherever rapid lookup matters. - **Compound interest.** Balance after n years at rate r: \( A_n = P(1+r)^n \). - **Algorithm complexity.** When computer scientists say an algorithm runs in \( O(n^2) \), they mean the operation count is a quadratic sequence in the input size. - **Population biology.** A population doubling every generation is geometric: \( P_n = P_0 \cdot 2^n \). - **Spectroscopy and physics.** The energy levels of a hydrogen atom follow \( E_n = -\frac{13.6}{n^2} \). The destination of an explicit formula is the ability to _jump_ — to predict what happens at the 1000th step without grinding through 999 of them. ## Where Students Lose Marks On Explicit Formulas ### **Mistake 1: Using n where you should use n−1** **Where it slips in:** Arithmetic and geometric sequence formulas. **Don't do this:** Write \( a_n = a_1 + n \) or \( a_n = a_1 \cdot r^n \). ### **Mistake 2: Confusing the recursive formula with the explicit formula** **Don't do this:** Write \( a_n = a_{n-1} + 4 \) when the question asks for the explicit form. ### **Mistake 3: Forgetting to verify the formula on the given terms** **Don't do this:** Derive the formula, write it down, move on. The five-bullet summary: - An **explicit formula** gives the n-th term \( a_n \) directly as a function of n. - Common mistake is the **off-by-one error** using n instead of n-1 in exponent or multiplier. - Binet's formula is the most famous closed form for Fibonacci's sequence. - Explicit formulas are essential wherever fast lookup matters.