Metamath Proof Explorer


Theorem nicomachus

Description: Nicomachus's Theorem. The sum of the odd numbers from N ^ 2 - N + 1 to N ^ 2 + N - 1 is N ^ 3 . Proof 2 from https://proofwiki.org/wiki/Nicomachus%27s_Theorem . (Contributed by SN, 21-Mar-2025)

Ref Expression
Assertion nicomachus ⊢ N ∈ ℕ 0 → ∑ k = 1 N N 2 − N + 2 ⁢ k - 1 = N 3

Proof

Step Hyp Ref Expression
1 fzfid ⊢ N ∈ ℕ 0 → 1 … N ∈ Fin
2 nn0cn ⊢ N ∈ ℕ 0 → N ∈ ℂ
3 2 adantr ⊢ N ∈ ℕ 0 ∧ k ∈ 1 … N → N ∈ ℂ
4 3 sqcld ⊢ N ∈ ℕ 0 ∧ k ∈ 1 … N → N 2 ∈ ℂ
5 4 3 subcld ⊢ N ∈ ℕ 0 ∧ k ∈ 1 … N → N 2 − N ∈ ℂ
6 2cnd ⊢ N ∈ ℕ 0 ∧ k ∈ 1 … N → 2 ∈ ℂ
7 elfznn ⊢ k ∈ 1 … N → k ∈ ℕ
8 7 nncnd ⊢ k ∈ 1 … N → k ∈ ℂ
9 8 adantl ⊢ N ∈ ℕ 0 ∧ k ∈ 1 … N → k ∈ ℂ
10 6 9 mulcld ⊢ N ∈ ℕ 0 ∧ k ∈ 1 … N → 2 ⁢ k ∈ ℂ
11 1cnd ⊢ N ∈ ℕ 0 ∧ k ∈ 1 … N → 1 ∈ ℂ
12 10 11 subcld ⊢ N ∈ ℕ 0 ∧ k ∈ 1 … N → 2 ⁢ k − 1 ∈ ℂ
13 1 5 12 fsumadd ⊢ N ∈ ℕ 0 → ∑ k = 1 N N 2 − N + 2 ⁢ k - 1 = ∑ k = 1 N N 2 − N + ∑ k = 1 N 2 ⁢ k − 1
14 id ⊢ N ∈ ℕ 0 → N ∈ ℕ 0
15 2 sqcld ⊢ N ∈ ℕ 0 → N 2 ∈ ℂ
16 15 2 subcld ⊢ N ∈ ℕ 0 → N 2 − N ∈ ℂ
17 14 16 fz1sumconst ⊢ N ∈ ℕ 0 → ∑ k = 1 N N 2 − N = N ⁢ N 2 − N
18 2 15 2 subdid ⊢ N ∈ ℕ 0 → N ⁢ N 2 − N = N ⁢ N 2 − N ⋅ N
19 df-3 ⊢ 3 = 2 + 1
20 19 oveq2i ⊢ N 3 = N 2 + 1
21 2nn0 ⊢ 2 ∈ ℕ 0
22 21 a1i ⊢ N ∈ ℕ 0 → 2 ∈ ℕ 0
23 2 22 expp1d ⊢ N ∈ ℕ 0 → N 2 + 1 = N 2 ⋅ N
24 20 23 eqtrid ⊢ N ∈ ℕ 0 → N 3 = N 2 ⋅ N
25 15 2 mulcomd ⊢ N ∈ ℕ 0 → N 2 ⋅ N = N ⁢ N 2
26 24 25 eqtr2d ⊢ N ∈ ℕ 0 → N ⁢ N 2 = N 3
27 2 sqvald ⊢ N ∈ ℕ 0 → N 2 = N ⋅ N
28 27 eqcomd ⊢ N ∈ ℕ 0 → N ⋅ N = N 2
29 26 28 oveq12d ⊢ N ∈ ℕ 0 → N ⁢ N 2 − N ⋅ N = N 3 − N 2
30 17 18 29 3eqtrd ⊢ N ∈ ℕ 0 → ∑ k = 1 N N 2 − N = N 3 − N 2
31 oddnumth ⊢ N ∈ ℕ 0 → ∑ k = 1 N 2 ⁢ k − 1 = N 2
32 30 31 oveq12d ⊢ N ∈ ℕ 0 → ∑ k = 1 N N 2 − N + ∑ k = 1 N 2 ⁢ k − 1 = N 3 - N 2 + N 2
33 3nn0 ⊢ 3 ∈ ℕ 0
34 33 a1i ⊢ N ∈ ℕ 0 → 3 ∈ ℕ 0
35 2 34 expcld ⊢ N ∈ ℕ 0 → N 3 ∈ ℂ
36 35 15 npcand ⊢ N ∈ ℕ 0 → N 3 - N 2 + N 2 = N 3
37 13 32 36 3eqtrd ⊢ N ∈ ℕ 0 → ∑ k = 1 N N 2 − N + 2 ⁢ k - 1 = N 3