**The Binomial Theorem 二項式定理**

**The Binomial Theorem**

(4.2)

**Proof of the Binomial Theorem by Induction:**

When *n*=1, Equation (4.2) reduces to

Assume Equation (4.2) for *n*-1. Now,

Letting *i*=*k*+1 in the first sum and *i*=*k* in the second sum, we find that

where the next to last equality follows by Equation (4.1). By induction the theorem is now proved.

**Combinatorial Proof of the Binomial Theorem:**

Consider the product
.
Its expansion consists
of the sum of 2^{n} terms, each term being the product of *n* factors. Further,
each of the 2^{n} terms in the sum will contain as a factor either *x*_{i} or
*y*_{i} for each
*i*=1,2,...,*n*. For example,
(*x*_{1}+*y*_{1})(*x*_{1}+*y*_{1})=*x*_{1}*x*_{2}+*x*_{1}*y*_{2}+*y*_{1}*x*_{2}+*y*_{1}*y*_{2} Now, how many of the 2^{n}terms in the sum will have as factors *k* of the *x*_{i}'s and (*n*-*k*) of the
*y*_{i}'s? As each term consisting of *k* of the *x*_{i}'s and (*n*-*k*) of the
*y*_{i}'s corresponds to a choice of a group of *k* from the *n* values
*x*_{1},*x*_{2},...,*x*_{n}, there are
such terms. Thus, letting
*x*_{i}=*x*, *y*_{i}=*y*, *i*=1,...,*n*, we see that