Arithmetic Series: Summing Arithmetic Sequences

An arithmetic series adds up an arithmetic sequence. And there's a beautiful shortcut.

What is 1 + 2 + 3 + 4 + ... + 100?

The slow way: add all 100 numbers.

The fast way: notice that 1 + 100 = 101. So does 2 + 99. And 3 + 98. There are 50 such pairs.

50 × 101 = 5,050.

This is the trick. Pair the first term with the last, the second with the second-to-last. Each pair has the same sum. Multiply by the number of pairs.

Sₙ = n(a₁ + aₙ)/2

That's the formula. The sum of n terms equals n times the average of the first and last term.

Why the Formula Works

An arithmetic sequence has terms: a₁, a₁ + d, a₁ + 2d, ..., a₁ + (n-1)d.

Write the sum forward: S = a₁ + (a₁ + d) + (a₁ + 2d) + ... + aₙ

Write it backward: S = aₙ + (aₙ - d) + (aₙ - 2d) + ... + a₁

Add these two equations, term by term: 2S = (a₁ + aₙ) + (a₁ + aₙ) + (a₁ + aₙ) + ... (n times)

Every pair sums to a₁ + aₙ. There are n pairs.

2S = n(a₁ + aₙ)

S = n(a₁ + aₙ)/2

Gauss reportedly discovered this as a schoolboy when his teacher assigned adding 1 to 100. The story may be legend, but the insight is real: pairing opposites reveals structure.

The Formula in Three Forms

The same formula appears in different disguises:

Form 1: First and last terms Sₙ = n(a₁ + aₙ)/2

Use when you know the first term, last term, and number of terms.

Form 2: First term and common difference

Since aₙ = a₁ + (n-1)d:

Sₙ = n[2a₁ + (n-1)d]/2

Use when you know the first term, common difference, and number of terms.

Form 3: Average times count

Sₙ = n × (average of first and last)

The arithmetic mean of the endpoints, multiplied by how many terms.

All three are the same formula rearranged.

Examples

Sum of 1 + 2 + 3 + ... + n

Here a₁ = 1, aₙ = n, so: Sₙ = n(1 + n)/2 = n(n+1)/2

This is the triangular number formula. The 100th triangular number is 100(101)/2 = 5,050.

Sum of first 20 odd numbers

Odd numbers: 1, 3, 5, 7, ..., 39 (the 20th odd number is 2(20) - 1 = 39)

a₁ = 1, a₂₀ = 39, n = 20

S₂₀ = 20(1 + 39)/2 = 20 × 20 = 400

Remarkable: the sum of the first n odd numbers always equals n². Check: 1 = 1², 1+3 = 4 = 2², 1+3+5 = 9 = 3².

Sum of 5 + 10 + 15 + ... + 100

a₁ = 5, aₙ = 100, d = 5

First, how many terms? From aₙ = a₁ + (n-1)d: 100 = 5 + (n-1)(5) 95 = 5(n-1) n = 20

S₂₀ = 20(5 + 100)/2 = 20 × 52.5 = 1,050

Sum of 100 + 97 + 94 + ... + 4

Decreasing sequence! d = -3, a₁ = 100, aₙ = 4

Number of terms: 4 = 100 + (n-1)(-3) -96 = -3(n-1) n = 33

S₃₃ = 33(100 + 4)/2 = 33 × 52 = 1,716

The formula works for decreasing sequences too.

Applications

Triangular stacking: If you stack objects in a triangle—1 on top, 2 in the next row, 3 in the next—the total for n rows is n(n+1)/2.

Handshakes: If n people each shake hands with everyone else, the total number of handshakes is n(n-1)/2 (everyone shakes with n-1 others, but each handshake is counted twice).

Stadium seating: If a stadium section has 20 seats in the first row and adds 2 seats per row for 30 rows, the total seats are:

• First row: 20, last row: 20 + 29(2) = 78
• Total: 30(20 + 78)/2 = 30 × 49 = 1,470 seats

Gauss's classroom: The original application. Add 1 to 100 in seconds, not minutes.

When Things Go Wrong

Mistake 1: Forgetting (n-1)

The nth term is a₁ + (n-1)d, not a₁ + nd. From first to nth is n-1 steps.

Mistake 2: Using the formula on non-arithmetic sequences

The formula only works when the difference between consecutive terms is constant. 1, 2, 4, 8 is not arithmetic—it's geometric. Different formula.

Mistake 3: Miscounting terms

How many integers from 5 to 15 inclusive? Not 15 - 5 = 10. It's 15 - 5 + 1 = 11.

General rule: from a to b inclusive has b - a + 1 terms.

Partial Sums and Infinite Series

For an arithmetic sequence, the partial sums Sₙ grow without bound (unless d = 0).

If d > 0: terms increase, sum grows faster and faster. If d < 0: terms eventually become negative, but the sum still diverges.

Arithmetic series never converge (except the trivial case where every term is 0). You can't add up an infinite arithmetic sequence and get a finite answer.

This is fundamentally different from geometric series, where convergence is possible when |r| < 1.

The Sum as a Quadratic

Notice that Sₙ = n[2a₁ + (n-1)d]/2 = (d/2)n² + (a₁ - d/2)n.

This is a quadratic in n. The graph of Sₙ vs. n is a parabola.

• If d > 0: parabola opens upward
• If d < 0: parabola opens downward

This explains why arithmetic series grow or shrink at an accelerating rate. The sum doesn't just increase linearly—it curves.

Connection to Integration

The sum 1 + 2 + 3 + ... + n ≈ n²/2 for large n.

Compare to the integral: ∫₀ⁿ x dx = n²/2.

This isn't coincidence. Summing integers is the discrete analog of integrating x. The sum formula and the integral formula are the same.

This connection deepens in calculus: Riemann sums (finite sums) approximate integrals (continuous sums). Arithmetic series are the simplest case.

Summary

The arithmetic series formula reduces tedious addition to simple arithmetic:

Sₙ = n(a₁ + aₙ)/2 = n[2a₁ + (n-1)d]/2

It works by pairing first with last, second with second-to-last, exploiting the symmetry of arithmetic growth.

The formula transforms O(n) additions into O(1) calculation. That's not just convenience—it's the power of recognizing pattern.

Part 6 of the Sequences Series series.

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