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 AN0A+1NAN=k=0N1(Nk)Ak

Proof

Step Hyp Ref Expression
1 fzfid AN00N1Fin
2 fzssp1 0N10N-1+1
3 nn0cn N0N
4 3 adantl AN0N
5 ax-1cn 1
6 npcan N1N-1+1=N
7 4 5 6 sylancl AN0N-1+1=N
8 7 oveq2d AN00N-1+1=0N
9 2 8 sseqtrid AN00N10N
10 9 sselda AN0k0N1k0N
11 bccl2 k0N(Nk)
12 11 adantl AN0k0N(Nk)
13 12 nncnd AN0k0N(Nk)
14 simpl AN0A
15 elfznn0 k0Nk0
16 expcl Ak0Ak
17 14 15 16 syl2an AN0k0NAk
18 13 17 mulcld AN0k0N(Nk)Ak
19 10 18 syldan AN0k0N1(Nk)Ak
20 1 19 fsumcl AN0k=0N1(Nk)Ak
21 expcl AN0AN
22 addcom A1A+1=1+A
23 14 5 22 sylancl AN0A+1=1+A
24 23 oveq1d AN0A+1N=1+AN
25 binom1p AN01+AN=k=0N(Nk)Ak
26 simpr AN0N0
27 nn0uz 0=0
28 26 27 eleqtrdi AN0N0
29 oveq2 k=N(Nk)=(NN)
30 oveq2 k=NAk=AN
31 29 30 oveq12d k=N(Nk)Ak=(NN)AN
32 28 18 31 fsumm1 AN0k=0N(Nk)Ak=k=0N1(Nk)Ak+(NN)AN
33 bcnn N0(NN)=1
34 33 adantl AN0(NN)=1
35 34 oveq1d AN0(NN)AN=1AN
36 21 mullidd AN01AN=AN
37 35 36 eqtrd AN0(NN)AN=AN
38 37 oveq2d AN0k=0N1(Nk)Ak+(NN)AN=k=0N1(Nk)Ak+AN
39 32 38 eqtrd AN0k=0N(Nk)Ak=k=0N1(Nk)Ak+AN
40 24 25 39 3eqtrd AN0A+1N=k=0N1(Nk)Ak+AN
41 20 21 40 mvrraddd AN0A+1NAN=k=0N1(Nk)Ak