Metamath Proof Explorer


Theorem bcp1m1

Description: Compute the binomial coefficient of ( N + 1 ) over ( N - 1 ) (Contributed by Scott Fenton, 11-May-2014) (Revised by Mario Carneiro, 22-May-2014)

Ref Expression
Assertion bcp1m1 ⊢ N ∈ ℕ 0 → ( N + 1 N − 1 ) = N + 1 ⋅ N 2

Proof

Step Hyp Ref Expression
1 peano2nn0 ⊢ N ∈ ℕ 0 → N + 1 ∈ ℕ 0
2 nn0z ⊢ N ∈ ℕ 0 → N ∈ ℤ
3 peano2zm ⊢ N ∈ ℤ → N − 1 ∈ ℤ
4 2 3 syl ⊢ N ∈ ℕ 0 → N − 1 ∈ ℤ
5 bccmpl ⊢ N + 1 ∈ ℕ 0 ∧ N − 1 ∈ ℤ → ( N + 1 N − 1 ) = ( N + 1 N + 1 - N − 1 )
6 1 4 5 syl2anc ⊢ N ∈ ℕ 0 → ( N + 1 N − 1 ) = ( N + 1 N + 1 - N − 1 )
7 nn0cn ⊢ N ∈ ℕ 0 → N ∈ ℂ
8 1cnd ⊢ N ∈ ℕ 0 → 1 ∈ ℂ
9 7 8 8 pnncand ⊢ N ∈ ℕ 0 → N + 1 - N − 1 = 1 + 1
10 df-2 ⊢ 2 = 1 + 1
11 9 10 eqtr4di ⊢ N ∈ ℕ 0 → N + 1 - N − 1 = 2
12 11 oveq2d ⊢ N ∈ ℕ 0 → ( N + 1 N + 1 - N − 1 ) = ( N + 1 2 )
13 bcn2 ⊢ N + 1 ∈ ℕ 0 → ( N + 1 2 ) = N + 1 ⁢ N + 1 - 1 2
14 1 13 syl ⊢ N ∈ ℕ 0 → ( N + 1 2 ) = N + 1 ⁢ N + 1 - 1 2
15 ax-1cn ⊢ 1 ∈ ℂ
16 pncan ⊢ N ∈ ℂ ∧ 1 ∈ ℂ → N + 1 - 1 = N
17 7 15 16 sylancl ⊢ N ∈ ℕ 0 → N + 1 - 1 = N
18 17 oveq2d ⊢ N ∈ ℕ 0 → N + 1 ⁢ N + 1 - 1 = N + 1 ⋅ N
19 18 oveq1d ⊢ N ∈ ℕ 0 → N + 1 ⁢ N + 1 - 1 2 = N + 1 ⋅ N 2
20 14 19 eqtrd ⊢ N ∈ ℕ 0 → ( N + 1 2 ) = N + 1 ⋅ N 2
21 12 20 eqtrd ⊢ N ∈ ℕ 0 → ( N + 1 N + 1 - N − 1 ) = N + 1 ⋅ N 2
22 6 21 eqtrd ⊢ N ∈ ℕ 0 → ( N + 1 N − 1 ) = N + 1 ⋅ N 2