# Calcolatore di successioni aritmetiche, geometriche e di Fibonacci

> The nth term and the sum of the first n terms of an arithmetic or geometric sequence, the sum to infinity, and exact Fibonacci numbers, with a table.

Versione interattiva: https://www.calcopenly.com/it/math/sequence-calculator
Argomento: Calcolatori di matematica

An arithmetic sequence adds a fixed difference d at each step, so aₙ = a₁ + (n − 1)d and the first n terms sum to Sₙ = n(a₁ + aₙ)/2. A geometric sequence multiplies by a fixed ratio r, so aₙ = a₁rⁿ⁻¹ and Sₙ = a₁(1 − rⁿ)/(1 − r); when |r| < 1 the sum to infinity is a₁/(1 − r). Fibonacci numbers start from F₀ = 0 and F₁ = 1, add the previous two, and are computed exactly to every digit.

Savings that grow by a fixed amount, compound growth and exam questions all use these formulas. The default, 3, 7, 11, …, reaches 39 at term 10, and those 10 terms sum to 210; the halving series 1 + 1/2 + 1/4 + … totals 1.998046875 after 10 terms and approaches 2.

A geometric series with |r| ≥ 1 has no finite sum to infinity, and an arithmetic series has one only when every term is 0. The Fibonacci sum F₁ + … + Fₙ equals Fₙ₊₂ − 1.

## Dati

- **Sequence** (opzioni: Aritmetica, Geometric, Fibonacci)
- **First term a₁**
- **Common difference d**
- **Common ratio r**
- **Term number n**: Fibonacci counts from F₀ = 0, F₁ = 1; the others from a₁.

## Risultati

- nth term — risultato principale
- Sum of the first n terms
- Sum to infinity
- nth term, all digits
- Digits in the term

## Formula

$$
\begin{gathered} a_n = a_1 + (n-1)d \\ S_n = \tfrac{n}{2}(a_1 + a_n) \\[10pt] a_n = a_1 r^{n-1} \\ S_n = a_1\frac{1 - r^n}{1 - r},\quad S_\infty = \frac{a_1}{1 - r} \\[10pt] F_n = F_{n-1} + F_{n-2} \end{gathered}
$$

## Esempi svolti

### Arithmetic 3, 7, 11, … to 10 terms

- Sequence: Aritmetica
- First term a₁: 3
- Common difference d: 4
- Term number n: 10
- **nth term: 39**
- **Sum of the first n terms: 210**
- Fonte di verifica: Python 3.8: sum(3 + 4k for k in range(10)) = 210, last term 39

### Arithmetic with a negative step

- Sequence: Aritmetica
- First term a₁: 100
- Common difference d: -7.5
- Term number n: 20
- **nth term: -42.5**
- **Sum of the first n terms: 575**
- Fonte di verifica: Python 3.8 decimal loop over 20 terms (forkA/verify_seq.py)

### Geometric 2, 6, 18, … to 6 terms

- Sequence: Geometric
- First term a₁: 2
- Common ratio r: 3
- Term number n: 6
- **nth term: 486**
- **Sum of the first n terms: 728**
- Fonte di verifica: Python 3.8: 2·3⁵ = 486, Σ 2·3^k (k < 6) = 728

### Halving series 1 + 1/2 + 1/4 + …

- Sequence: Geometric
- First term a₁: 1
- Common ratio r: 0.5
- Term number n: 10
- **nth term: 0.001953125**
- **Sum of the first n terms: 1.998046875**
- **Sum to infinity: 2**
- Fonte di verifica: Python fractions: Σ (1/2)^k for k < 10 = 1023/512; sum to infinity 1/(1 − 1/2) = 2

### Ratio 1 (edge)

- Sequence: Geometric
- First term a₁: 5
- Common ratio r: 1
- Term number n: 4
- **nth term: 5**
- **Sum of the first n terms: 20**
- Fonte di verifica: With r = 1 every term is a₁, so Sₙ = n·a₁ = 20

### Fibonacci F₁₀₀

- Sequence: Fibonacci
- Term number n: 100
- **nth term: 3.54225 × 10²⁰**
- **nth term, all digits: 354224848179261915075**
- **Digits in the term: 21**
- **Sum of the first n terms: 9.27373 × 10²⁰**
- Fonte di verifica: OEIS A000045 (F₁₀₀); sum F₁₀₂ − 1 from a Python 3.8 loop

## Domande

### How do you find the nth term of an arithmetic sequence?

Use aₙ = a₁ + (n − 1)d: start from the first term and add the common difference n − 1 times. For 3, 7, 11, … the difference is 4, so the 10th term is 3 + 9 × 4 = 39 and the 50th is 3 + 49 × 4 = 199. To find d from two terms, divide their difference by the gap in positions: (39 − 3)/(10 − 1) = 4.

### What is the formula for the sum of an arithmetic series?

Sₙ = n(a₁ + aₙ)/2, the number of terms times the average of the first and last term. For 3 + 7 + … + 39, which has 10 terms, S = 10 × (3 + 39)/2 = 210. The pairing idea behind it is often credited to the young Gauss, who summed 1 to 100 as 50 pairs of 101 to get 5,050.

### How do you find the sum of a geometric series?

For n terms, Sₙ = a₁(1 − rⁿ)/(1 − r) whenever r ≠ 1. For 2 + 6 + 18 + … with 6 terms, S = 2 × (1 − 3⁶)/(1 − 3) = 2 × (−728)/(−2) = 728. When r = 1 every term equals a₁, so Sₙ = n × a₁ and 5 + 5 + 5 + 5 = 20.

### When does a geometric series have a sum to infinity?

Only when the common ratio is strictly between −1 and 1. Then rⁿ shrinks toward 0 and the sum approaches a₁/(1 − r): 1 + 1/2 + 1/4 + … = 1/(1 − 1/2) = 2, and 0.9 + 0.09 + 0.009 + … = 0.9/(1 − 0.1) = 1, which is why 0.999… equals 1. With |r| ≥ 1 the terms do not shrink, so the series grows without bound or oscillates.

### What is the 100th Fibonacci number?

F₁₀₀ = 354,224,848,179,261,915,075, a 21-digit number, counting from F₀ = 0 and F₁ = 1 as in OEIS A000045. Consecutive Fibonacci numbers grow by a factor approaching the golden ratio φ ≈ 1.618034, so Fₙ is close to φⁿ/√5 and each term adds about 0.209 digits.

### Quanto è preciso «Calcolatore di successioni aritmetiche, geometriche e di Fibonacci»?

La precisione dipende dai dati inseriti e dalle ipotesi del metodo. Il calcolo decimale usa 50 cifre significative, ma stime, metodi numerici e dati di origine possono essere meno precisi; l’arrotondamento visualizzato non elimina questi limiti. Esempi svolti verificati con fonti indipendenti: 7. Per esempio, «Arithmetic 3, 7, 11, … to 10 terms» viene verificato con Python 3.8: sum(3 + 4k for k in range(10)) = 210, last term 39.

### Da dove proviene il metodo?

Abramowitz & Stegun, Handbook of Mathematical Functions, §3.6 (series); OEIS A000045 — Fibonacci numbers; Wikipedia — Geometric series.

## Fonti

- Abramowitz & Stegun, Handbook of Mathematical Functions, §3.6 (series)
- [OEIS A000045 — Fibonacci numbers](https://oeis.org/A000045)
- [Wikipedia — Geometric series](https://en.wikipedia.org/wiki/Geometric_series)