Metamath Proof Explorer


Theorem binom11

Description: Special case of the binomial theorem for 2 ^ N . (Contributed by Mario Carneiro, 13-Mar-2014)

Ref Expression
Assertion binom11 ⊢ N ∈ ℕ 0 → 2 N = ∑ k = 0 N ( N k)

Proof

Step Hyp Ref Expression
1 df-2 ⊢ 2 = 1 + 1
2 1 oveq1i ⊢ 2 N = 1 + 1 N
3 ax-1cn ⊢ 1 ∈ ℂ
4 binom1p ⊢ 1 ∈ ℂ ∧ N ∈ ℕ 0 → 1 + 1 N = ∑ k = 0 N ( N k) ⁢ 1 k
5 3 4 mpan ⊢ N ∈ ℕ 0 → 1 + 1 N = ∑ k = 0 N ( N k) ⁢ 1 k
6 2 5 eqtrid ⊢ N ∈ ℕ 0 → 2 N = ∑ k = 0 N ( N k) ⁢ 1 k
7 elfzelz ⊢ k ∈ 0 … N → k ∈ ℤ
8 1exp ⊢ k ∈ ℤ → 1 k = 1
9 7 8 syl ⊢ k ∈ 0 … N → 1 k = 1
10 9 oveq2d ⊢ k ∈ 0 … N → ( N k) ⁢ 1 k = ( N k) ⋅ 1
11 bccl2 ⊢ k ∈ 0 … N → ( N k) ∈ ℕ
12 11 nncnd ⊢ k ∈ 0 … N → ( N k) ∈ ℂ
13 12 mulridd ⊢ k ∈ 0 … N → ( N k) ⋅ 1 = ( N k)
14 10 13 eqtrd ⊢ k ∈ 0 … N → ( N k) ⁢ 1 k = ( N k)
15 14 sumeq2i ⊢ ∑ k = 0 N ( N k) ⁢ 1 k = ∑ k = 0 N ( N k)
16 6 15 eqtrdi ⊢ N ∈ ℕ 0 → 2 N = ∑ k = 0 N ( N k)