Binomial Theorem: Let a0 , a1 , a2 ,
an1 , an be the nth row of Pascal’s triangle. Then
( x  y ) n  a0 x n  a1 x n 1 y  a2 x n  2 y 2  an 1 x1 y n 1  an y n for each integer n  0 .
Proof (by Induction):
1. Base Case: The 0th row of Pascals’ triangle is: 1. Since ( x  y)0  1  1x0 , the theorem
holds for n  0 .
2. Inductive step: Suppose that for some integer n  0 , we have that
( x  y ) n  a0 x n  a1 x n 1 y  a2 x n  2 y 2  an 1 x1 y n 1  an y n , where a0 , a1 , a2 ,
an1 , an is the
nth row of Pascal’s triangle. We first note that the (n  1) st row would then be
a0 , a0  a1 , a1  a2 ,
, an1  an , an .
We then have that
( x  y ) n 1  ( x  y ) n  ( x  y )
 (a0 x n  a1 x n 1 y  a2 x n  2 y 2 
 a0 x n 1  a1 x n y  a2 x n 1 y 2 
 a0 x n y  a1 x n 1 y 2 
 an 1 x1 y n 1  an y n )( x  y )
 an 1 x 2 y n 1  an xy n
(multipling each term by x )
 an  2 x 2 y n 1  an 1 xy n  an y n 1
 a0 x n 1  (a0  a1 ) x n y  (a1  a2 ) x n 1 y 2 
(multipling each term by y )
 (an 1  an ) xy n  an y n 1
Therefore, the coefficients of the terms in the expansion of ( x  y )n 1 are simply the
entries in the (n  1) st row of Pascal’s triangle and we have proved the theorem with the
loving use of the Principle of Mathematical Induction.