Metamath Proof Explorer


Theorem binom1dif

Description: A summation for the difference between ( ( A + 1 ) ^ N ) and ( A ^ N ) . (Contributed by Scott Fenton, 9-Apr-2014) (Revised by Mario Carneiro, 22-May-2014)

Ref Expression
Assertion binom1dif A N 0 A + 1 N A N = k = 0 N 1 ( N k) A k

Proof

Step Hyp Ref Expression
1 fzfid A N 0 0 N 1 Fin
2 fzssp1 0 N 1 0 N - 1 + 1
3 nn0cn N 0 N
4 3 adantl A N 0 N
5 ax-1cn 1
6 npcan N 1 N - 1 + 1 = N
7 4 5 6 sylancl A N 0 N - 1 + 1 = N
8 7 oveq2d A N 0 0 N - 1 + 1 = 0 N
9 2 8 sseqtrid A N 0 0 N 1 0 N
10 9 sselda A N 0 k 0 N 1 k 0 N
11 bccl2 k 0 N ( N k)
12 11 adantl A N 0 k 0 N ( N k)
13 12 nncnd A N 0 k 0 N ( N k)
14 simpl A N 0 A
15 elfznn0 k 0 N k 0
16 expcl A k 0 A k
17 14 15 16 syl2an A N 0 k 0 N A k
18 13 17 mulcld A N 0 k 0 N ( N k) A k
19 10 18 syldan A N 0 k 0 N 1 ( N k) A k
20 1 19 fsumcl A N 0 k = 0 N 1 ( N k) A k
21 expcl A N 0 A N
22 addcom A 1 A + 1 = 1 + A
23 14 5 22 sylancl A N 0 A + 1 = 1 + A
24 23 oveq1d A N 0 A + 1 N = 1 + A N
25 binom1p A N 0 1 + A N = k = 0 N ( N k) A k
26 simpr A N 0 N 0
27 nn0uz 0 = 0
28 26 27 eleqtrdi A N 0 N 0
29 oveq2 k = N ( N k) = ( N N)
30 oveq2 k = N A k = A N
31 29 30 oveq12d k = N ( N k) A k = ( N N) A N
32 28 18 31 fsumm1 A N 0 k = 0 N ( N k) A k = k = 0 N 1 ( N k) A k + ( N N) A N
33 bcnn N 0 ( N N) = 1
34 33 adantl A N 0 ( N N) = 1
35 34 oveq1d A N 0 ( N N) A N = 1 A N
36 21 mulid2d A N 0 1 A N = A N
37 35 36 eqtrd A N 0 ( N N) A N = A N
38 37 oveq2d A N 0 k = 0 N 1 ( N k) A k + ( N N) A N = k = 0 N 1 ( N k) A k + A N
39 32 38 eqtrd A N 0 k = 0 N ( N k) A k = k = 0 N 1 ( N k) A k + A N
40 24 25 39 3eqtrd A N 0 A + 1 N = k = 0 N 1 ( N k) A k + A N
41 20 21 40 mvrraddd A N 0 A + 1 N A N = k = 0 N 1 ( N k) A k