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 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
4 binom1p 1 N 0 1 + 1 N = k = 0 N ( N k) 1 k
5 3 4 mpan N 0 1 + 1 N = k = 0 N ( N k) 1 k
6 2 5 syl5eq N 0 2 N = k = 0 N ( N k) 1 k
7 elfzelz k 0 N k
8 1exp k 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)
12 11 nncnd k 0 N ( N k)
13 12 mulid1d 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 0 2 N = k = 0 N ( N k)