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等差、等比与斐波那契数列计算器

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.

更新于 已验证的示例：7

Fibonacci counts from F₀ = 0, F₁ = 1; the others from a₁.
试一试
nth term
nth term: 39
最大小数位数：10；取最近值，等距时远离零
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
条款 行数：10
kTerm aₖRunning sum Sₖ
133
2710
31121
41536
51955
62378
727105
831136
935171
1039210
计算方法 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.

关于等差、等比与斐波那契数列计算器

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.

计算示例

Arithmetic 3, 7, 11, … to 10 terms

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

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

Arithmetic with a negative step

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

核验来源：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

核验来源：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

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

常见问题

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.

“等差、等比与斐波那契数列计算器”有多准确？

准确性取决于输入值和方法的假设。十进制运算使用50位有效数字，但估算、数值方法和源数据的精度可能较低；显示时的舍入并不能消除这些限制。 已按独立来源核验的计算示例：7。 例如，“Arithmetic 3, 7, 11, … to 10 terms”根据Python 3.8: sum(3 + 4k for k in range(10)) = 210, last term 39进行核验。

这种方法出自哪里？

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

关于此计算器

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}

来源

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

已对照来源验证

此计算器包含 7 个已解示例，答案来自独立来源。这些示例会在测试套件中运行，你也可以在此运行验证。

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