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The Binomial Theorem.

$$\begin {align*} \left( x + y \right)^n &= \displaystyle\sum_{k=0}^{n} \dbinom {n}{k} x^{n-k} y^k \\&= \dbinom {n}{0} x^n y^0 + \dbinom {n}{1} x^{n-1} y^1 + \cdots + \dbinom {n}{n-1} x^1 y^{n-1} + \dbinom {n}{n} x^0 y^n. \end {align*}$$

Proof. We use mathematical induction on $n$. When $n=0$, both sides equal $1$, and it is easy to see that the statement is true. For the inductive step, say we are given that the statement is true for some $n-1$. We will now prove it for $n$. $$\begin {align*} \left( x + y \right)^{n} &= \left( x + y \right) \cdot \left( x + y \right)^{n-1} \\&= \left( x + y \right) \cdot \left[ \displaystyle\sum_{k=0}^{n-1} \dbinom {n-1}{k} x^{(n-1)-k} y^k \right] \\&= \displaystyle\sum_{k=0}^{n-1} \dbinom {n-1}{k} x^{n-k} y^k + \displaystyle\sum_{j=0}^{n-1} \dbinom {n-1}{j} x^{(n-1)-j} y^{j+1} \\&= \displaystyle\sum_{k=0}^{n-1} \dbinom {n-1}{k} x^{n-k} y^k + \displaystyle\sum_{k=1}^{n} \dbinom {n-1}{k-1} x^{n-k} y^k \\&= \displaystyle\sum_{k=0}^{n} \left[ \dbinom {n-1}{k} + \dbinom {n-1}{k-1} \right] \cdot x^{n-k} y^k \\&= \displaystyle\sum_{k=0}^{n} \dbinom {n}{k} x^{n-k} y^k, \end {align*}$$ as desired. $\blacksquare$

The Binomial Theorem states that when $n$ is a positive integer, the binomial expansion of $(a+b)^n$ is $${n \choose 0}a^nb^0+{n \choose 1}a^{n-1}b^1+{n \choose 2}a^{n-2}b^2+{n \choose 3}a^{n-3}b^3+\cdots+{n \choose n}a^0b^n.$$

This has $n+1$ terms.

• The notation for a combination (sometimes read as "$n$ choose $k$"): $${n \choose k} = \frac{n!}{k!(n-k)!}$$

• The value of ${n \choose k}$ can also be found using the Pascal's Triangle: it is the $k$-th number from left to right in the $n$-th row. (Remember that the Pascal's Triangle begins with the $0$-th row and number!)

Another way to evaluate the Choose Notation is to multiply $n$ by $(n-1)$, then $(n-2)$, and so on, for k times. Then divide the quantity by the factorial of $k,$ or $k!$.

(Video) Applications of the Binomial Theorem

by NotesCollegeAlgebra

In this video by NotesCollegeAlgebra and application to the binomial theorem for approximating an exponentiation of a decimal is worked out.

The normal approximation to the binomial distribution

Note: You should read other explanations about binomial theorem, properties of binomial coefficient first.

Here is an application on how to use binomial theorem to find normal approximation to the binomial distribution.

A coin is flipped $n$ times with probability of getting heads is $p$. This is a binomial distribution $X \sim B(n,p)$. The Year 12 MATH B textbook (Australia) gives an normal approximation to this: $B(n,p) \sim N(np,np(1-p))=N(\mu, \sigma^2)$ where $\mu=E[X]$ is the mean (or expected value) and $\sigma$ is the standard deviation. Since the textbook doesn't give a proof for this so I will go and prove. :D

According to the definition, the expected value (or the mean) of getting heads is $E[X]=\mu= \sum_{i=0}^n i\binom{n}{i}p^i(1-p)^{n-i}$. It suffices to prove that $\sum_{i=0}^n i\binom{n}{i}p^i(1-p)^{n-i} =np$. We have $$\begin{aligned} \frac{1}{np}\sum_{i=0}^n i\binom{n}{i}p^i(1-p)^{n-i} & = \sum_{i=1}^n\frac in \binom{n}{i} p^{i-1}(1-p)^{n-i}, \\ & = \sum_{i=1}^n\binom{n-1}{i-1} p^{i-1}(1-p)^{n-i}, \\ & = \sum_{i=0}^{n-1}\binom{n-1}{i}p^i(1-p)^{n-1-i} = 1. \end{aligned}$$ Now, for $\sigma^2$, according to the definition, we have $$\sigma^2= \sum_{i=0}^n(i-\mu)^2 \binom ni p^i(1-p)^{n-i}= \sum_{i=0}^ni^2 \binom ni p^i(1-p)^{n-i}-n^2p^2.$$ Hence, it suffices to prove that $$\sum_{i=0}^ni^2 \binom ni p^i(1-p)^{n-i}= np(1+np-p).$$ Indeed, we have $$\begin{aligned} \frac{1}{np}\sum_{i=0}^ni^2 \binom ni p^i(1-p)^{n-i} & = \sum_{i=1}^ni \binom{n-1}{i-1}p^{i-1}(1-p)^{n-i}, \\ & = \sum_{i=0}^{n-1}i\binom{n-1}{i}p^i(1-p)^{n-i-1}\\&+\sum_{i=0}^{n-1}\binom{n-1}{i}p^{i}(1-p)^{n-i-1}, \\ & = (n-1)p+1. \end{aligned}$$ We are done. $\square$

The Binomial Theorem

Right now, we have been multiplying out factors in order to get expanded polynomials. However, as we get to binomials with larger and larger degrees, it will become way too much work to multiply out all of the factors. That's where the Binomial Theorem comes in! We can use the solved term in the expansion.

Now, we can take a shortcut when expanding binomials. We can use this formula to expand binomial expressions with really large powers. Think about it: if you had to expand $(x+3)^19$, do you really want to multiply out all of those factors?

No! We can use the binomial coefficients to make our lives simpler. This formula also gives us a nifty cheat, called Pascal's Triangle. Once you expand a binomial, they follow the same pattern.

Looking Further

We can also use factorial notation to expand polynomials, which you will learn about in the next lesson. You may also want to review the factorial.