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.

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.