PROOF 36

Prove.

The sum of an arithmetic sequence

On real sequences.

A way to read this proof

First separate what is assumed from what is to be proved. At each step, ask which definition or earlier result permits the move. If it is unclear, return to that premise rather than skipping to the conclusion.

Definition D1

The sequence (aₙ) satisfies aₙ₊₁ = aₙ + d with first term a₁ and common difference d.

Proof.

This equation determines aₙ uniquely for each natural number n. d does not vary with n.convention

∎

Definition D2

Sₙ is the sum from a₁ to aₙ, defined by S₁ = a₁, Sₖ₊₁ = Sₖ + aₖ₊₁.

Proof.

These two equations are used in place of an ellipsis. An ellipsis is an omission, not a rule.convention

∎

Definition D3

a ÷ b means the x satisfying b × x = a.

Proof.

a ÷ b has a value only when such an x is unique.convention

∎

Lemma L1

For every natural number n, aₙ₊₁ = a₁ + nd

Proof.

(1)Let Q(n) be “aₙ₊₁ = a₁ + nd”.convention

(2)a₂ = a₁ + dD1

(3)1 × d = d × 1 = dA5, A6

(4)a₂ = a₁ + 1×d, so Q(1) holds.(2), (3)

(5)Suppose Q(k). aₖ₊₁ = a₁ + kdassumption

(6)aₖ₊₂ = aₖ₊₁ + dD1

(7)　　 = (a₁ + kd) + d(5), (6)

(8)　　 = a₁ + (kd + d)(7), A2

(9)(k+1)d = d(k+1) = dk + d×1 = kd + dA5, A4, A6

(10)aₖ₊₂ = a₁ + (k+1)d, so Q(k+1) holds.(8), (9)

(11)Therefore Q(n) holds for every n.(4), (10), A1

∎

Lemma L2

Let P(n) be “2Sₙ = n(a₁ + aₙ)”. Then P(1) holds.

Proof.

(1)S₁ = a₁D2

(2)2S₁ = 2a₁(1), E

(3)2a₁ = (1+1)a₁ = a₁(1+1) = a₁×1 + a₁×1 = a₁ + a₁A7, A5, A4, A6

(4)1(a₁+a₁) = (a₁+a₁)×1 = a₁ + a₁A5, A6

(5)2S₁ = 1(a₁+a₁)(2), (3), (4)

∎

Lemma L3

For every natural number k, if P(k), then P(k+1).

Proof.

(1)Suppose 2Sₖ = k(a₁ + aₖ).assumption

(2)(k+1)(a₁+aₖ₊₁) = k(a₁+aₖ₊₁) + (a₁+aₖ₊₁)A5, A4, A6

(3)k(a₁+aₖ₊₁) = ka₁ + kaₖ₊₁A4

(4)aₖ₊₁ = aₖ + dD1

(5)kaₖ₊₁ = k(aₖ + d) = kaₖ + kd(4), E, A4

(6)(k+1)(a₁+aₖ₊₁) = (ka₁ + kaₖ) + (a₁ + kd) + aₖ₊₁(2), (3), (5), A2, A3

(7)ka₁ + kaₖ = k(a₁ + aₖ)A4

(8)a₁ + kd = aₖ₊₁L1

(9)aₖ₊₁ + aₖ₊₁ = aₖ₊₁(1+1) = aₖ₊₁ × 2 = 2aₖ₊₁A6, A4, A7, A5

(10)(k+1)(a₁+aₖ₊₁) = k(a₁+aₖ) + 2aₖ₊₁(6), (7), (8), (9), A2

(11)　　　　　　　 = 2Sₖ + 2aₖ₊₁(1), (10)

(12)2Sₖ + 2aₖ₊₁ = 2(Sₖ + aₖ₊₁) = 2Sₖ₊₁A4, D2

(13)2Sₖ₊₁ = (k+1)(a₁+aₖ₊₁), so P(k+1) holds.(11), (12)

∎

Theorem T1

For every natural number n, 2Sₙ = n(a₁ + aₙ)

Proof.

(1)P(1) holds.L2

(2)For every k, P(k) ⟹ P(k+1).L3

(3)Therefore P(n) holds for every n.(1), (2), A1

∎

Corollary C

For every natural number n, Sₙ = n(a₁ + aₙ)/2

Proof.

(1)2Sₙ = n(a₁ + aₙ)T1

(2)n(a₁+aₙ) ÷ 2 is the x with 2x = n(a₁+aₙ).D3

(3)Sₙ is such an x.(1), (2)

(4)If 2x = 2y then x = y, so such an x is unique.A8

(5)Sₙ = n(a₁ + aₙ)/2(3), (4), D3

∎

Remark

Can A1 be removed from this proof?

This proof contains no picture of adding the sum in reverse. Check what that picture rests on as well.