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)

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 2timesd ⊢ N ∈ ℕ → 2 ⋅ N = N + N
25 24 oveq2d ⊢ N ∈ ℕ → N 2 − 2 ⋅ N = N 2 − N + N
26 23 sqcld ⊢ N ∈ ℕ → N 2 ∈ ℂ
27 26 23 23 subsub4d ⊢ N ∈ ℕ → N 2 - N - N = N 2 − N + N
28 25 27 eqtr4d ⊢ N ∈ ℕ → N 2 − 2 ⋅ N = N 2 - N - N
29 28 oveq1d ⊢ N ∈ ℕ → N 2 - 2 ⋅ N + 1 = N 2 − N - N + 1
30 binom2sub1 ⊢ N ∈ ℂ → N − 1 2 = N 2 - 2 ⋅ N + 1
31 23 30 syl ⊢ N ∈ ℕ → N − 1 2 = N 2 - 2 ⋅ N + 1
32 26 23 subcld ⊢ N ∈ ℕ → N 2 − N ∈ ℂ
33 1cnd ⊢ N ∈ ℕ → 1 ∈ ℂ
34 32 23 33 subsubd ⊢ N ∈ ℕ → N 2 - N - N − 1 = N 2 − N - N + 1
35 29 31 34 3eqtr4d ⊢ N ∈ ℕ → N − 1 2 = N 2 - N - N − 1
36 35 oveq1d ⊢ N ∈ ℕ → N − 1 2 + N - 1 = N 2 - N - N − 1 + N - 1
37 ax-1cn ⊢ 1 ∈ ℂ
38 subcl ⊢ N ∈ ℂ ∧ 1 ∈ ℂ → N − 1 ∈ ℂ
39 23 37 38 sylancl ⊢ N ∈ ℕ → N − 1 ∈ ℂ
40 32 39 npcand ⊢ N ∈ ℕ → N 2 - N - N − 1 + N - 1 = N 2 − N
41 36 40 eqtrd ⊢ N ∈ ℕ → N − 1 2 + N - 1 = N 2 − N
42 41 oveq1d ⊢ N ∈ ℕ → N − 1 2 + N - 1 2 = N 2 − N 2
43 22 42 eqtrd ⊢ N ∈ ℕ → ∑ k = 1 N − 1 k = N 2 − N 2
44 20 43 eqtrd ⊢ N ∈ ℕ → 0 + ∑ k = 0 + 1 N − 1 k = N 2 − N 2
45 9 44 eqtrd ⊢ N ∈ ℕ → ∑ k = 0 N − 1 k = N 2 − N 2
46 oveq1 ⊢ N = 0 → N − 1 = 0 − 1
47 46 oveq2d ⊢ N = 0 → 0 … N − 1 = 0 … 0 − 1
48 0re ⊢ 0 ∈ ℝ
49 ltm1 ⊢ 0 ∈ ℝ → 0 − 1 < 0
50 48 49 ax-mp ⊢ 0 − 1 < 0
51 0z ⊢ 0 ∈ ℤ
52 peano2zm ⊢ 0 ∈ ℤ → 0 − 1 ∈ ℤ
53 51 52 ax-mp ⊢ 0 − 1 ∈ ℤ
54 fzn ⊢ 0 ∈ ℤ ∧ 0 − 1 ∈ ℤ → 0 − 1 < 0 ↔ 0 … 0 − 1 = ∅
55 51 53 54 mp2an ⊢ 0 − 1 < 0 ↔ 0 … 0 − 1 = ∅
56 50 55 mpbi ⊢ 0 … 0 − 1 = ∅
57 47 56 eqtrdi ⊢ N = 0 → 0 … N − 1 = ∅
58 57 sumeq1d ⊢ N = 0 → ∑ k = 0 N − 1 k = ∑ k ∈ ∅ k
59 sum0 ⊢ ∑ k ∈ ∅ k = 0
60 58 59 eqtrdi ⊢ N = 0 → ∑ k = 0 N − 1 k = 0
61 sq0i ⊢ N = 0 → N 2 = 0
62 id ⊢ N = 0 → N = 0
63 61 62 oveq12d ⊢ N = 0 → N 2 − N = 0 − 0
64 0m0e0 ⊢ 0 − 0 = 0
65 63 64 eqtrdi ⊢ N = 0 → N 2 − N = 0
66 65 oveq1d ⊢ N = 0 → N 2 − N 2 = 0 2
67 2cn ⊢ 2 ∈ ℂ
68 2ne0 ⊢ 2 ≠ 0
69 67 68 div0i ⊢ 0 2 = 0
70 66 69 eqtrdi ⊢ N = 0 → N 2 − N 2 = 0
71 60 70 eqtr4d ⊢ N = 0 → ∑ k = 0 N − 1 k = N 2 − N 2
72 45 71 jaoi ⊢ N ∈ ℕ ∨ N = 0 → ∑ k = 0 N − 1 k = N 2 − N 2
73 1 72 sylbi ⊢ N ∈ ℕ 0 → ∑ k = 0 N − 1 k = N 2 − N 2