Calculus 2nd Edition Rogawski

APPENDIX C INDUCTION AND THE BINOMIAL THEOREM A15
( ( n n
= = 1. In summation notation,
0)
n)
(a + b) n =
n∑
k=0
( n
k)
a k b n−k
Pascal’s Triangle (described in the marginal note on page A14) can be used to compute
binomial coefficients if n and k are not too large. The Binomial Theorem provides the
following general formula:
( n n! n(n − 1)(n − 2) ···(n − k + 1)
=
=
k)
k! (n − k)! k(k − 1)(k − 2) ···2 · 1
Before proving this formula, we prove a recursion relation for binomial coefficients. Note,
however, that Eq. (3) is certainly correct for k = 0 and k = n (recall that by convention,
0! =1):
( n
=
0)
n!
(n − 0)! 0! = n! ( ) n
n! = 1, =
n
n!
(n − n)! n! = n!
n! = 1
3
THEOREM2 Recursion Relation for Binomial Coefficients
( ( n n − 1
=
k)
k
)
+
( ) n − 1
k − 1
for 1 ≤ k ≤ n − 1
Proof We write (a + b) n as (a + b)(a + b) n−1 and expand in terms of binomial coefficients:
(a + b) n = (a + b)(a + b) n−1
n∑
( n ∑n−1
( ) n − 1
a
k)
n−k b k = (a + b)
a n−1−k b k
k
k=0
k=0
∑n−1
( ) n − 1
∑n−1
( ) n − 1
= a
a n−1−k b k + b
a n−1−k b k
k
k
k=0
k=0
∑n−1
( ) n − 1 ∑n−1
( ) n − 1
=
a n−k b k +
a n−(k+1) b k+1
k
k
k=0
Replacing k by k − 1 in the second sum, we obtain
k=0
n∑
k=0
( n
k)
a n−k b k =
∑n−1
( n − 1
k=0
k
k=0
)
a n−k b k +
n∑
k=1
( ) n − 1
a n−k b k
k − 1
On the right-hand side, the first term in the first sum is a n and the last term in the second
sum is b n . Thus, we have
n∑
( (
n ∑n−1
(( ) ( )) )
n − 1 n − 1
a
k)
n−k b k = a n +
+ a n−k b k + b n
k k − 1
k=1
The recursion relation follows because the coefficients of a n−k b k on the two sides of the
equation must be equal.