Section 2.6 The Binomial Theorem
Here is a truly basic result from combinatorics kindergarten.
Theorem 2.30. Binomial Theorem.
Let \(x\) and \(y\) be real numbers with \(x\text{,}\) \(y\) and \(x+y\) non-zero. Then for every non-negative integer \(n\text{,}\)
\begin{equation*}
(x+y)^n=\sum_{i=0}^{n}\binom{n}{i}x^{n-i}y^{i}.
\end{equation*}
Proof.
View \((x+y)^n\) as a product
\begin{equation*}
(x+y)^n=\underbrace{(x+y)(x+y)(x+y)(x+y)\dots(x+y)(x+y)}_{n\text{ factors} }.
\end{equation*}
Each term of the expansion of the product results from choosing either \(x\) or \(y\) from one of these factors. If \(x\) is chosen \(n-i\) times and \(y\) is chosen \(i\) times, then the resulting product is \(x^{n-i}y^i\text{.}\) Clearly, the number of such terms is \(C(n,i)\text{,}\) i.e., out of the \(n\) factors, we choose the element \(y\) from \(i\) of them, while we take \(x\) in the remaining \(n-i\text{.}\)
Example 2.31.
There are times when we are interested not in the full expansion of a power of a binomial, but just the coefficient on one of the terms. The
Binomial Theorem gives that the coefficient of
\(x^5y^8\) in
\((2x-3y)^{13}\) is
\(\binom{13}{5}2^{5}(-3)^8\text{.}\)
Reading Questions Reading Questions
1.
Explain how you can use
Binomial Theorem to give yet another proof of the identity
\begin{equation*}
2^n = \binom{n}{0} + \binom{n}{1} + \cdots + \binom{n}{n-1}+\binom{n}{n} = \sum_{k=0}^n \binom{n}{k}\text{.}
\end{equation*}
You have attempted
of
activities on this page.
