Metamath Proof Explorer


Theorem arisum2

Description: Arithmetic series sum of the first N nonnegative integers. (Contributed by Mario Carneiro, 17-Apr-2015) (Proof shortened by AV, 2-Aug-2021) (Proof shortened by Timothy J. O'Malley, 9-Oct-2026)

Ref Expression
Assertion arisum2 ⊢ N ∈ ℕ 0 → ∑ k = 0 N − 1 k = N 2 − N 2

Proof

Step Hyp Ref Expression
1 elnn0 ⊢ N ∈ ℕ 0 ↔ N ∈ ℕ ∨ N = 0
2 nnm1nn0 ⊢ N ∈ ℕ → N − 1 ∈ ℕ 0
3 nn0uz ⊢ ℕ 0 = ℤ ≥ 0
4 2 3 eleqtrdi ⊢ N ∈ ℕ → N − 1 ∈ ℤ ≥ 0
5 elfznn0 ⊢ k ∈ 0 … N − 1 → k ∈ ℕ 0
6 5 adantl ⊢ N ∈ ℕ ∧ k ∈ 0 … N − 1 → k ∈ ℕ 0
7 6 nn0cnd ⊢ N ∈ ℕ ∧ k ∈ 0 … N − 1 → k ∈ ℂ
8 id ⊢ k = 0 → k = 0
9 4 7 8 fsum1p ⊢ N ∈ ℕ → ∑ k = 0 N − 1 k = 0 + ∑ k = 0 + 1 N − 1 k
10 1e0p1 ⊢ 1 = 0 + 1
11 10 oveq1i ⊢ 1 … N − 1 = 0 + 1 … N − 1
12 11 sumeq1i ⊢ ∑ k = 1 N − 1 k = ∑ k = 0 + 1 N − 1 k
13 12 oveq2i ⊢ 0 + ∑ k = 1 N − 1 k = 0 + ∑ k = 0 + 1 N − 1 k
14 fzfid ⊢ N ∈ ℕ → 1 … N − 1 ∈ Fin
15 elfznn ⊢ k ∈ 1 … N − 1 → k ∈ ℕ
16 15 adantl ⊢ N ∈ ℕ ∧ k ∈ 1 … N − 1 → k ∈ ℕ
17 16 nncnd ⊢ N ∈ ℕ ∧ k ∈ 1 … N − 1 → k ∈ ℂ
18 14 17 fsumcl ⊢ N ∈ ℕ → ∑ k = 1 N − 1 k ∈ ℂ
19 18 addlidd ⊢ N ∈ ℕ → 0 + ∑ k = 1 N − 1 k = ∑ k = 1 N − 1 k
20 13 19 eqtr3id ⊢ N ∈ ℕ → 0 + ∑ k = 0 + 1 N − 1 k = ∑ k = 1 N − 1 k
21 arisum ⊢ N − 1 ∈ ℕ 0 → ∑ k = 1 N − 1 k = N − 1 2 + N - 1 2
22 2 21 syl ⊢ N ∈ ℕ → ∑ k = 1 N − 1 k = N − 1 2 + N - 1 2
23 nncn ⊢ N ∈ ℕ → N ∈ ℂ
24 23 sqcld ⊢ N ∈ ℕ → N 2 ∈ ℂ
25 24 23 subcld ⊢ N ∈ ℕ → N 2 − N ∈ ℂ
26 2 nn0cnd ⊢ N ∈ ℕ → N − 1 ∈ ℂ
27 23 2timesd ⊢ N ∈ ℕ → 2 ⋅ N = N + N
28 27 oveq2d ⊢ N ∈ ℕ → N 2 − 2 ⋅ N = N 2 − N + N
29 24 23 23 subsub4d ⊢ N ∈ ℕ → N 2 - N - N = N 2 − N + N
30 28 29 eqtr4d ⊢ N ∈ ℕ → N 2 − 2 ⋅ N = N 2 - N - N
31 30 oveq1d ⊢ N ∈ ℕ → N 2 - 2 ⋅ N + 1 = N 2 − N - N + 1
32 binom2sub1 ⊢ N ∈ ℂ → N − 1 2 = N 2 - 2 ⋅ N + 1
33 23 32 syl ⊢ N ∈ ℕ → N − 1 2 = N 2 - 2 ⋅ N + 1
34 1cnd ⊢ N ∈ ℕ → 1 ∈ ℂ
35 25 23 34 subsubd ⊢ N ∈ ℕ → N 2 - N - N − 1 = N 2 − N - N + 1
36 31 33 35 3eqtr4d ⊢ N ∈ ℕ → N − 1 2 = N 2 - N - N − 1
37 25 26 36 mvrrsubd ⊢ N ∈ ℕ → N − 1 2 + N - 1 = N 2 − N
38 37 oveq1d ⊢ N ∈ ℕ → N − 1 2 + N - 1 2 = N 2 − N 2
39 22 38 eqtrd ⊢ N ∈ ℕ → ∑ k = 1 N − 1 k = N 2 − N 2
40 9 20 39 3eqtrd ⊢ N ∈ ℕ → ∑ k = 0 N − 1 k = N 2 − N 2
41 oveq1 ⊢ N = 0 → N − 1 = 0 − 1
42 41 oveq2d ⊢ N = 0 → 0 … N − 1 = 0 … 0 − 1
43 fz00m1 ⊢ 0 … 0 − 1 = ∅
44 42 43 eqtrdi ⊢ N = 0 → 0 … N − 1 = ∅
45 44 sumeq1d ⊢ N = 0 → ∑ k = 0 N − 1 k = ∑ k ∈ ∅ k
46 sum0 ⊢ ∑ k ∈ ∅ k = 0
47 45 46 eqtrdi ⊢ N = 0 → ∑ k = 0 N − 1 k = 0
48 sq0i ⊢ N = 0 → N 2 = 0
49 id ⊢ N = 0 → N = 0
50 48 49 oveq12d ⊢ N = 0 → N 2 − N = 0 − 0
51 0m0e0 ⊢ 0 − 0 = 0
52 50 51 eqtrdi ⊢ N = 0 → N 2 − N = 0
53 52 oveq1d ⊢ N = 0 → N 2 − N 2 = 0 2
54 2cn ⊢ 2 ∈ ℂ
55 2ne0 ⊢ 2 ≠ 0
56 54 55 div0i ⊢ 0 2 = 0
57 53 56 eqtrdi ⊢ N = 0 → N 2 − N 2 = 0
58 47 57 eqtr4d ⊢ N = 0 → ∑ k = 0 N − 1 k = N 2 − N 2
59 40 58 jaoi ⊢ N ∈ ℕ ∨ N = 0 → ∑ k = 0 N − 1 k = N 2 − N 2
60 1 59 sylbi ⊢ N ∈ ℕ 0 → ∑ k = 0 N − 1 k = N 2 − N 2