Induction step of the proof of the binomial theorem
Assume for some arbitrary natural number m, that for arbitrary a and b:
m 
m 
m 
m 
(a  b) m   a m   a m1b1   a m2b2 ...  b m
0 
1 
2 
m 
Consider case m+1. We need to show the binomial equation works for this case.

Expand the left side of
(a  b) m1 by the distributive law:
(a  b) m1  (a  b)(a  b) m 
m 
m 
m 
m  
a a m  a m1b1   a m2 b 2  ...  b m 
1 
2 
m  
0 
m 
m 
m 
m  
b a m   a m1b1   a m2 b 2  ...  b m 
1 
2 
m  
0 
by using the induction hypothesis twice. Now multiply through each term of the first row by a and each term of the second row
by b:


m  m1 m  m 1 m  m1 2
m  m 
a

a
b

a
b

...
 
 
 
 ab 
0
1
2
 
 
m 
 

m  m 1 m  m1 2 m  m2 3
m  m1 
a
b

a
b

a
b

...
 
 
 
 b 
0
1
2
 
 
m 
 


Then carefully combine similar terms (same powers of a and b). Notice each term (after the first) is similar to the term below
and to the left. The first and last terms have nothing similar. So:

m   m  m m  m1
m  m1 m  m  m 1 m  m  m1 2
a


a
b


a
b

...
   
   
  
 
ab   b
1
0
2
1
m
m
1
0 












m 





(Almost done.) Now notice the first and last binomial coefficients are 1. We could shift the m’s to m+1 and they are still 1.


And then – notice Pascal’s rule allows us to rewrite each of the sums of 2 binomial coefficients above. So we get exactly what
we want-the right side:

m 1 m1 m 1 m 1 m 1 m1 2
m 1 m m 1 m1
a

a
b

a
b

...







ab  
b
 0 
 1 
 2 
 m 
m 1