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Aritmetik, geometrik ve Fibonacci dizisi hesaplayıcı

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.

Güncellendi Doğrulanan örnekler: 7

Fibonacci counts from F₀ = 0, F₁ = 1; the others from a₁.
Dene
nth term
nth term: 39
En fazla ondalık basamak: 10; En yakına; eşit uzaklıkta sıfırdan uzağa
Sum of the first n terms
210

Starting at 3 and adding 4 each time, term 10 is 39 and the first 10 terms add up to 210.

Terms and running sum

050100150200246810Term number kValue
Term aₖRunning sum
Koşullar Satır sayısı: 10
kTerm aₖRunning sum Sₖ
133
2710
31121
41536
51955
62378
727105
831136
935171
1039210
Nasıl hesaplanır S

  1. nth term

    an=a1+(n−1)d=3+(10−1)×4=39a_n = a_1 + (n - 1)d = 3 + (10 - 1) \times 4 = 39

  2. Sum of the first n terms

    Sn=n2(a1+an)=102(3+39)=210S_n = \frac{n}{2}(a_1 + a_n) = \frac{10}{2}\left(3 + 39\right) = 210

    Pair the first and last terms: each pair has the same total.

  3. Sum to infinity

    An arithmetic series with a non-zero term or step has no finite sum to infinity.

Aritmetik, geometrik ve Fibonacci dizisi hesaplayıcı hakkında

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.

Çözümlü örnekler

Arithmetic 3, 7, 11, … to 10 terms

Sequence
Aritmetik
First term a₁
3
Common difference d
4
Term number n
10
nth term
39
Sum of the first n terms
210

Doğrulama kaynağı: Python 3.8: sum(3 + 4k for k in range(10)) = 210, last term 39

Arithmetic with a negative step

Sequence
Aritmetik
First term a₁
100
Common difference d
-7.5
Term number n
20
nth term
-42.5
Sum of the first n terms
575

Doğrulama kaynağı: 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

Doğrulama kaynağı: 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

Doğrulama kaynağı: Python fractions: Σ (1/2)^k for k < 10 = 1023/512; sum to infinity 1/(1 − 1/2) = 2

Sorular

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.

“Aritmetik, geometrik ve Fibonacci dizisi hesaplayıcı” ne kadar doğru sonuç verir?

Doğruluk, girdilerinize ve yöntemin varsayımlarına bağlıdır. Ondalık aritmetik 50 anlamlı basamak kullanır; ancak tahminler, sayısal yöntemler ve kaynak veriler daha az hassas olabilir. Gösterilen değerin yuvarlanması bu sınırları ortadan kaldırmaz. Bağımsız kaynaklarla doğrulanan çözümlü örnek sayısı: 7. Örneğin “Arithmetic 3, 7, 11, … to 10 terms”, Python 3.8: sum(3 + 4k for k in range(10)) = 210, last term 39 ile karşılaştırılarak doğrulanır.

Yöntemin kaynağı nedir?

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

Bu hesaplayıcı hakkında

an=a1+(n−1)dSn=n2(a1+an)an=a1rn−1Sn=a11−rn1−r,S∞=a11−rFn=Fn−1+Fn−2\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}

Kaynaklar

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

Kaynaklarla doğrulandı

Bu hesaplayıcı, yanıtları bağımsız kaynaklardan alınan 7 çözümlü örnek içerir. Bunlar test paketinde çalıştırılır; burada da çalıştırabilirsiniz.

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