Metamath Proof Explorer


Theorem binom21

Description: Special case of binom2 where B = 1 . (Contributed by Scott Fenton, 11-May-2014)

Ref Expression
Assertion binom21 ⊢ A ∈ ℂ → A + 1 2 = A 2 + 2 ⁢ A + 1

Proof

Step Hyp Ref Expression
1 ax-1cn ⊢ 1 ∈ ℂ
2 binom2 ⊢ A ∈ ℂ ∧ 1 ∈ ℂ → A + 1 2 = A 2 + 2 ⁢ A ⋅ 1 + 1 2
3 1 2 mpan2 ⊢ A ∈ ℂ → A + 1 2 = A 2 + 2 ⁢ A ⋅ 1 + 1 2
4 mulrid ⊢ A ∈ ℂ → A ⋅ 1 = A
5 4 oveq2d ⊢ A ∈ ℂ → 2 ⁢ A ⋅ 1 = 2 ⁢ A
6 5 oveq2d ⊢ A ∈ ℂ → A 2 + 2 ⁢ A ⋅ 1 = A 2 + 2 ⁢ A
7 sq1 ⊢ 1 2 = 1
8 7 a1i ⊢ A ∈ ℂ → 1 2 = 1
9 6 8 oveq12d ⊢ A ∈ ℂ → A 2 + 2 ⁢ A ⋅ 1 + 1 2 = A 2 + 2 ⁢ A + 1
10 3 9 eqtrd ⊢ A ∈ ℂ → A + 1 2 = A 2 + 2 ⁢ A + 1