# Binomial Theorem

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The Binomial Theorem – HMC Calculus Tutorial

We know that \begin{eqnarray*} (x+y)^0&=&1\\ (x+y)^1&=&x+y\\ (x+y)^2&=&x^2+2xy+y^2 \end{eqnarray*} and we can easily expand $(x+y)^3=x^3+3x^2y+3xy^2+y^3.$ For higher powers, the expansion gets very tedious by hand! Fortunately, the Binomial Theorem gives us the expansion for any positive integer power of $(x+y)$:

For any positive integer $n$, $(x+y)^n=\sum^n_{k=0} \left(\begin{array}{c} n\\ k \end{array}\right)x^{n-k}y^k$ where $$\displaystyle\left(\begin{array}{c} n\\ k \end{array}\right)=\frac{(n)(n-1)(n-2)\cdots(n-(k-1))}{k!}=\frac{n!}{k!(n-k)!}.$$

Binomial Theorem, Proof by Induction
Binomial Theorem

For any positive integer $n$, $(x+y)^n=\sum^n_{k=0} \left(\begin{array}{c} n\\ k \end{array}\right)x^{n-k}y^k.$

#### Proof by Induction:

For $n=1$, $(x+y)^1=x+y=\left(\begin{array}{c} 1\\ 0 \end{array}\right)x^{1-0}y^0+\left(\begin{array}{c} 1\\ 1 \end{array}\right)x^{1-1}y^1=\sum_{k=0}^1 \left(\begin{array}{c} 1\\ k \end{array}\right)x^{1-k}y^k.$

Suppose $\displaystyle (x+y)^{n-1}=\sum^{n-1}_{k=0} \left(\begin{array}{c} n-1\\ k \end{array}\right)x^{(n-1)-k}y^k.$

Consider $(x+y)^n$. \begin{eqnarray*} (x+y)^n&=&(x+y)(x+y)^{n-1}\\ &=&(x+y)\left[\sum^{n-1}_{k=0} \left(\begin{array}{c} n-1\\ k \end{array}\right)x^{(n-1)-k}y^k\right]\\ &=& \sum^{n-1}_{k=0} \left(\begin{array}{c} n-1\\ k \end{array}\right)x^{n-k}y^k + \sum^{n-1}_{j=0} \left(\begin{array}{c} n-1\\ j \end{array}\right)x^{(n-1)-j}y^{j+1}\\ &=&\sum^{n-1}_{k=0} \left(\begin{array}{c} n-1\\ k \end{array}\right)x^{n-k}y^k+\sum^{n-1}_{j=0} \left(\begin{array}{c} n-1\\ (j+1)-1 \end{array}\right)x^{n-(j+1)}y^{j+1}\\ &=&\sum^{n-1}_{k=0} \left(\begin{array}{c} n-1\\ k \end{array}\right)x^{n-k}y^k+\sum^{n}_{k=1} \left(\begin{array}{c} n-1\\ k-1 \end{array}\right)x^{n-k}y^{k}\\ &=&\sum^{n}_{k=0} \left[\left(\begin{array}{c} n-1\\ k \end{array}\right)x^{n-k}y^k\right]-\left(\begin{array}{c} n-1\\ n \end{array}\right)x^0y^n\\ &~&+\sum^{n}_{k=0} \left[\left(\begin{array}{c} n-1\\ k-1 \end{array}\right)x^{n-k}y^{k}\right]-\left(\begin{array}{c} n-1\\ -1 \end{array}\right)x^ny^0\\ &=&\sum^{n}_{k=0} \left[\left(\begin{array}{c} n-1\\ k \end{array}\right)+\left(\begin{array}{c} n-1\\ k-1 \end{array}\right)\right]x^{n-k}y^k\\ &=& \sum^{n}_{k=0} \left(\begin{array}{c} n\\ k \end{array}\right)x^{n-k}y^k \end{eqnarray*} and the theorem is proved!

Binomial Theorem, Combinatorial Proof
Binomial Theorem

For any positive integer $n$, $(x+y)^n=\sum^n_{k=0} \left(\begin{array}{c} n\\ k \end{array}\right)x^{n-k}y^k.$

#### Combinatorial Proof:

Since $(x+y)^n=\underbrace{(x+y)(x+y)\cdots(x+y)}_{\textrm{n of these}}$ each term in $(x+y)^n$ has the form $x^{n-k}y^k$ for some $k$ between $0$ and $n$, inclusive.

The coefficient of $x^{n-k}y^k$ for a particular $k$ is just the number of ways to choose $k$ factors of $y$ from the $n$ factors of $(x+y)$, with factors of $x$ coming from the remaining $(n-k)$ factors. The number of ways to choose $k$ objects from a collection of $n$ objects, without replacement and order not important, is just $\displaystyle \left(\begin{array}{c} n\\ k \end{array}\right)$.

Thus, the term $x^{n-k}y^k$ has coefficient $\displaystyle \left(\begin{array}{c} n\\ k \end{array}\right)$, so $(x+y)^n=\sum^n_{k=0} \left(\begin{array}{c} n\\ k \end{array}\right)x^{n-k}y^k.$

Connection to Pascal’s Triangle
Connection to Pascal’s Triangle
$\begin{array}{ccc} (x+y)^0&=&1\\ (x+y)^1&=&1x+1y\\ (x+y)^2&=&1x^2+2xy+1y^2\\ (x+y)^3&=&1x^3+3x^2y+3xy^2+1y^3\\ (x+y)^4&=&1x^4+4x^3y+6x^2y^2+4xy^3+1y^4\\ (x+y)^5&=&1x^5+5x^4y+10x^3y^2+10x^2y^3+5xy^4+1y^5 \end{array}$

Do you see the connection??

###### Example

By the Binomial Theorem, \begin{eqnarray*} (x+y)^3&=& \sum_{k=0}^3 \left(\begin{array}{c} 3\\ k \end{array}\right)x^{3-k}y^k\\ &=& \left(\begin{array}{c} 3\\ 0 \end{array}\right)x^3+\left(\begin{array}{c} 3\\ 1 \end{array}\right)x^2y+\left(\begin{array}{c} 3\\ 2 \end{array}\right)xy^2+\left(\begin{array}{c} 3\\ 3 \end{array}\right)y^3\\ &=& x^3+3x^2y+3xy^2+y^3 \end{eqnarray*} as expected.

#### Extensions of the Binomial Theorem

A useful special case of the Binomial Theorem is $(1+x)^n=\sum^n_{k=0} \left(\begin{array}{c} n\\ k \end{array}\right)x^k$ for any positive integer $n$, which is just the Taylor series for $(1+x)^n$.

This formula can be extended to all real powers $\alpha$: $(1+x)^{\alpha}=\sum^{\infty}_{k=0} \left(\begin{array}{c} \alpha\\ k \end{array}\right)x^k$ for any real number $\alpha$, where $\left(\begin{array}{c} \alpha\\ k \end{array}\right)=\frac{(\alpha)(\alpha -1)(\alpha -2)\cdots (\alpha -(k-1))}{k!}=\frac{\alpha !}{k!(\alpha -k)!}.$ Notice that the formula now gives an infinite series: when $\alpha=n$ is a positive integer, all but the first $(n+1)$ terms are $0$ since after this $n-n$ ($=0$) appears in each numerator.

This expansion is very useful for approximating $(1+x)^{\alpha}$ for $|x|\ll 1$: $(1+x)^\alpha =1+\alpha x+ \frac{\alpha(\alpha -1)}{2!}x^2+\frac{\alpha (\alpha -1)(\alpha -2)}{3!}x^3+\cdots.$ But for $|x|\ll 1$, higher powers of $x$ get small very quickly, so $(1+x)^{\alpha}$ can be approximated to any accuracy we need by truncating the series after a finite number of terms.

###### Example

For $|x|\ll 1$, \begin{eqnarray*} (1+x)^{5/2}&\approx& 1+\frac{5}{2}x,\\ (1-2x)^{100}&\approx& 1-200x,\\ (1+x^2)^{-3}&\approx& 1-3x^2. \end{eqnarray*} This type of reasoning is useful in investigating what happens when a physical system is perturbed slightly, introducing a new very small term $x$.

#### Key Concepts

###### Binomial Theorem

For any positive integer $n$, $$(x+y)^n = \sum_{k=0}^n \left(\begin{array}{c} n \\ k \end{array}\right) x^{n-k} y^k$$ where $$\left(\begin{array}{c} n \\ k \end{array}\right) = \frac{n(n-1)(n-2)\cdots(n-k+1)}{k!} = \frac{n!}{k!(n-k)!}.$$