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 N02N=k=0N(Nk)

Proof

Step Hyp Ref Expression
1 df-2 2=1+1
2 1 oveq1i 2N=1+1N
3 ax-1cn 1
4 binom1p 1N01+1N=k=0N(Nk)1k
5 3 4 mpan N01+1N=k=0N(Nk)1k
6 2 5 eqtrid N02N=k=0N(Nk)1k
7 elfzelz k0Nk
8 1exp k1k=1
9 7 8 syl k0N1k=1
10 9 oveq2d k0N(Nk)1k=(Nk)1
11 bccl2 k0N(Nk)
12 11 nncnd k0N(Nk)
13 12 mulridd k0N(Nk)1=(Nk)
14 10 13 eqtrd k0N(Nk)1k=(Nk)
15 14 sumeq2i k=0N(Nk)1k=k=0N(Nk)
16 6 15 eqtrdi N02N=k=0N(Nk)