Arithmetic Series (Gauss Sum Formula)
The Gauss sum formula adds terms with constant difference: number of terms times the mean of first and last term.
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Formula
s_{n} = \frac{n \cdot (a_{1} + a_{n})}{2}Variables & units – Arithmetic Series (Gauss Sum Formula)
| Symbol | Meaning | Unit |
|---|---|---|
| sₙ | Sum of the first n terms | dimensionless |
| n | Number of terms | dimensionless |
| a₁ | First term of the series | dimensionless |
| aₙ | Last term: aₙ = a₁ + (n−1)·d | dimensionless |
Derivation & background – Arithmetic Series (Gauss Sum Formula)
Famous anecdote: nine-year-old Gauss added the numbers 1 to 100 in seconds by forming 50 pairs with equal sum 101 (1+100, 2+99, ...), hence 50·101 = 5050. Exactly this is inside the formula: couple first and last term, every pair sums equally. Special case 1 + 2 + ... + n = n(n+1)/2. With the constant difference d the variant sₙ = n/2·(2a₁ + (n−1)·d) applies.
Exam blueprint
Validity range
Holds for arithmetic sequences, i.e. constant difference d between the terms; the pair formula needs first and last term, the d-variant only a₁, d and n.
Derivation steps
Gauss trick: writing the sum forwards and backwards pairs terms with equal sum.
- 1sₙ = a₁ + a₂ + ... + aₙ and backwards sₙ = aₙ + aₙ₋₁ + ... + a₁; each of the n columns sums to a₁ + aₙ.
- 2So 2sₙ = n·(a₁ + aₙ), dividing by 2 gives the formula.
Rearrangements
Variant with difference d
When the last term is not given.
Little Gauss
Special case a₁ = 1, d = 1; the classic.
Last term
Needed first when only a₁, d and n are known.
Task variant
Compute 1 + 2 + ... + 100.
s₁₀₀ = 100·101/2 = 5050 (50 pairs with sum 101).
a₁ = 5, d = 3, n = 20: compute the sum s₂₀.
a₂₀ = 5 + 19·3 = 62, so s₂₀ = 20·(5 + 62)/2 = 20·33.5 = 670. Check with the d-variant: 10·(10 + 57) = 670 ✓.
Common mistakes
Determining the number of terms n incorrectly.
From a to b with step d there are n = (b − a)/d + 1 terms, not (b − a)/d.
Applying the formula to non-arithmetic sums.
Only with constant difference; 1 + 2 + 4 + 8 is geometric.
Shortcutting n(n+1)/2 to n²/2.
The +1 belongs there: 100·101/2 = 5050, not 5000.
Exam context
- Counting and sum tasks, linear instalment models, mathematical induction, runtime sums in computer science.
These mistakes cost points in real exams. The set drills them until they stick.
Formula cluster
Sequences and series
Additive counterpart of the geometric series; showcase for proof ideas.
Worked example
1 + 2 + ... + 100 = 100·101/2 = 5050. In general: a₁ = 5, d = 3, n = 20: a₂₀ = 5 + 19·3 = 62 and s₂₀ = 20·(5 + 62)/2 = 670.
Applications
Sum and counting tasks, seat row and stacking problems, linear depreciation and instalment plans, induction proofs, computer science (loop costs)
Quanta exam set
Curated exam set for "Arithmetic Series (Gauss Sum Formula)":
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Which formula describes Arithmetic Series (Gauss Sum Formula)?
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How do you rearrange sₙ = n·(a₁ + aₙ)/2 for Variant with difference d?
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Which common mistake happens with Arithmetic Series (Gauss Sum Formula)?
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Frequently asked questions about Arithmetic Series (Gauss Sum Formula)
How did Gauss add the numbers from 1 to 100 so quickly?+
According to the anecdote, nine-year-old Gauss's class was told to add the numbers 1 to 100, and he had the answer within seconds: he paired the first with the last term, the second with the second-to-last and so on: 1 + 100 = 101, 2 + 99 = 101, 3 + 98 = 101, ... That gives 50 pairs each summing to 101, hence 50·101 = 5050. Exactly this pairing sits inside the general formula sₙ = n·(a₁ + aₙ)/2: n terms form n/2 pairs with the constant sum a₁ + aₙ. The trick works for every arithmetic series, because the growth at the front and the shrinking at the back balance exactly.
Which variant of the sum formula do you use when?+
There are two equivalent forms. If you know the first AND the last term, take the pair form sₙ = n·(a₁ + aₙ)/2; it is the fastest. If instead you know a₁, the constant difference d and the count n, take sₙ = n/2·(2a₁ + (n−1)·d), or first compute aₙ = a₁ + (n−1)·d and then switch to the pair form. Example: a₁ = 5, d = 3, n = 20: aₙ = 5 + 19·3 = 62 and s₂₀ = 20·(5 + 62)/2 = 670; the d-form directly gives 10·(10 + 57) = 670 ✓. Both formulas are the same statement, once expressed with aₙ and once with d.
How do you determine the number of terms of a sum?+
With n = (aₙ − a₁)/d + 1: range divided by step width, plus 1. The "+1" is crucial and most often forgotten, because the starting term counts too. Examples: from 1 to 100 in steps of one there are (100 − 1)/1 + 1 = 100 terms. The even numbers from 2 to 100 have d = 2, so (100 − 2)/2 + 1 = 50 terms, and their sum is 50·(2 + 100)/2 = 2550. From 17 to 71 in steps of three: (71 − 17)/3 + 1 = 19 terms. Check: if the division does not come out evenly, the supposed final term does not belong to the sequence at all; then determine the largest fitting term first.
Which sum formulas hold for even and odd numbers?+
Both are arithmetic series with d = 2, and two neat short formulas arise. The first n even numbers: 2 + 4 + ... + 2n = n·(n + 1); example n = 50: 50·51 = 2550, identical to the pair formula 50·(2 + 100)/2 ✓. The first n odd numbers: 1 + 3 + ... + (2n − 1) = n²; example: 1 + 3 + 5 + 7 = 16 = 4². The second formula has a famous geometric interpretation: laying an L-shaped border of the next odd number of stones around a square grows it square by square. Both identities are standard exercises for mathematical induction and quick check values in exams.
What is the difference between an arithmetic and a geometric series?+
The building law of the terms. Arithmetic means: constant DIFFERENCE, each term arises by adding d (5, 8, 11, 14 with d = 3); the sum grows polynomially, formula sₙ = n·(a₁ + aₙ)/2. Geometric means: constant FACTOR, each term arises by multiplying by q (3, 6, 12, 24 with q = 2); for |q| > 1 the sum grows exponentially, formula sₙ = a₁·(qⁿ − 1)/(q − 1). Quick test on data: differences of neighbouring terms constant → arithmetic; ratios constant → geometric. Typical applications: linear instalments and seat rows arithmetic, compound interest and growth processes geometric. Only the geometric series can have a finite limit, for |q| < 1.
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How do you calculate with Arithmetic Series (Gauss Sum Formula)?
Here is how to work through a typical Arithmetic Series (Gauss Sum Formula) (sₙ = n·(a₁ + aₙ)/2) task step by step:
- 1
Task
Compute 1 + 2 + ... + 100.
Solution path
s₁₀₀ = 100·101/2 = 5050 (50 pairs with sum 101).
- 2
Task
a₁ = 5, d = 3, n = 20: compute the sum s₂₀.
Solution path
a₂₀ = 5 + 19·3 = 62, so s₂₀ = 20·(5 + 62)/2 = 20·33.5 = 670. Check with the d-variant: 10·(10 + 57) = 670 ✓.