## What is Binomial Theorem?

A binomial is a polynomial that is the sum of two terms, each of which is a monomial, i.e. single term. Consider a simple binomial (*x*+*y*), where *x* and *y* are two monomials.

${\left(x+y\right)}^{1}=x+y...\left(1\right)$

What is the result if this binomial is multiplied by itself, i.e. the square of a binomial?

$\left(x+y\right)\left(x+y\right)={x}^{2}+2xy+{y}^{2}...\left(2\right)$

What is the result of multiplying three binomials?

${\left(x+y\right)}^{3}={x}^{3}+3{x}^{2}y+3x{y}^{2}+{y}^{3}...\left(3\right)$

Notice the pattern when the brackets are expanded.

- In the first expression, if the binomial is raised to a power of 1, the result has two terms upon expansion.
- In the second expression, the binomial is raised to a power of 2, and the result has three terms upon expansion.
- In the third expression, the binomial is raised to a power of 3, and the result has four terms upon expansion.
- Therefore, if the binomial is raised to a power of
*n*, the result will have*n+1*number of terms.

If the binomial is multiplied with itself *n* times, it is impractical to multiply the brackets one after the other. Hence, we use the binomial formula given by the binomial theorem. It gives the expansion of the polynomial in the form of sum of terms involved in the expression (*x* and *y* in the case above).

## Binomial Functions and Formulas

The binomial formula is given as

${\left(a+b\right)}^{n}=\sum _{k=0}^{n}\left(\begin{array}{c}n\\ k\end{array}\right){a}^{n-k}{b}^{k}$

Where $\left(\begin{array}{c}n\\ k\end{array}\right)$ is read as *n choose k*. It means for a given set of *n*, in how many ways can you choose *k* elements from it. Mathematically it is interpreted as

The expression *n choose k* helps in getting the coefficients for the terms upon expansion.

Let us verify *n choose k *for Expression 1:

${\left(x+y\right)}^{1}=x+y...\left(1\right)$

For *n* = 1, the value of *k* will be 0, 1. There will be two terms in the expression, hence two coefficients as well.

$\left(\begin{array}{c}1\\ 0\end{array}\right)=\frac{1!}{\left(1-0\right)!0!}=1\phantom{\rule{0ex}{0ex}}\left(\begin{array}{c}1\\ 1\end{array}\right)=\frac{1!}{\left(1-1\right)!1!}=1$

The coefficients in the expansion for (*a*+*b*)^{1 }are 1 and 1 respectively.

Now, let’s verify *n choose k *for Expression 2:

$\left(x+y\right)\left(x+y\right)={x}^{2}+2xy+{y}^{2}...\left(2\right)$

For *n* = 2, the value of *k* will be 0, 1, 2. There will be three terms in the expressions, and therefore, three coefficients.

$\left(\begin{array}{c}2\\ 0\end{array}\right)=\frac{2!}{\left(2-0\right)!0!}=1\phantom{\rule{0ex}{0ex}}\left(\begin{array}{c}2\\ 1\end{array}\right)=\frac{2!}{\left(2-1\right)!1!}=2\phantom{\rule{0ex}{0ex}}\left(\begin{array}{c}2\\ 2\end{array}\right)=\frac{2!}{\left(2-2\right)!2!}=1$

We can see that in Expression 2, the coefficients are 1, 2, 1.

Now, for Expression 4, let’s use the binomial formula to evaluate the binomial raised to the power of 4:

${(x+y)}^{4}$

$=\sum _{k=0}^{4}\left(\begin{array}{c}4\\ k\end{array}\right){x}^{4-k}{y}^{k}$

$=\left(\begin{array}{c}4\\ 0\end{array}\right){\left(x\right)}^{4-0}{\left(y\right)}^{0}+\left(\begin{array}{c}4\\ 1\end{array}\right){\left(x\right)}^{4-1}{\left(y\right)}^{1}+\left(\begin{array}{c}4\\ 2\end{array}\right){\left(x\right)}^{4-2}{\left(y\right)}^{2}+\left(\begin{array}{c}4\\ 3\end{array}\right){\left(x\right)}^{4-3}{\left(y\right)}^{3}+\left(\begin{array}{c}4\\ 4\end{array}\right){\left(x\right)}^{4-4}\left(y\right)4$

$=\left(\begin{array}{c}4\\ 0\end{array}\right){x}^{4}+\left(\begin{array}{c}4\\ 1\end{array}\right){x}^{3}y+\left(\begin{array}{c}4\\ 2\end{array}\right){x}^{2}{y}^{2}+\left(\begin{array}{c}4\\ 3\end{array}\right)x{y}^{3}+\left(\begin{array}{c}4\\ 4\end{array}\right){y}^{4}$

$=\frac{4!}{(4-0)!0!}{x}^{4}+\frac{4!}{(4-1)!1!}{x}^{3}y+\frac{4!}{(4-2)!2!}{x}^{2}{y}^{2}+\frac{4!}{(4-3)!3!}x{y}^{3}+\frac{4!}{(4-4)!4!}{y}^{4}$

$={x}^{4}+4{x}^{3}y+6{x}^{2}{y}^{2}+4x{y}^{3}+{y}^{4}...\left(4\right)$

The coefficients obtained from the *n* choose *k* formula can also be obtained from a Pascal’s triangle.

The relation between the two can be seen here:

If Pascal’s triangle can give the desired results, then why do we need the *n choose k* formula?

To find the coefficient using Pascal’s triangle, we will have to build the triangle to include the given number of coefficients. If the given number is too big, the computation will be a laborious task. The process can be simplified using the *n choose k* formula.

## Binomials Wrap Up and More Samples

Let’s compare Expressions 1 – 4 below.

${(x+y)}^{1}={x}^{1}{y}^{0}+{x}^{0}{y}^{1}$

$=x+y$ ...(1)

${(x+y)}^{2}={x}^{2}{y}^{0}+2{x}^{1}{y}^{1}+{x}^{0}{y}^{2}$

$={x}^{2}+2xy+{y}^{2}$ ...(2)

${(x+y)}^{3}={x}^{3}{y}^{0}+3{x}^{2}{y}^{1}+3{x}^{1}{y}^{2}+{x}^{0}{y}^{3}$

$={x}^{3}+3{x}^{2}y+3x{y}^{2}+{y}^{3}$ ...(3)

${(x+y)}^{4}={x}^{4}{y}^{0}+4{x}^{3}{y}^{1}+6{x}^{2}{y}^{2}+4{x}^{1}{y}^{3}+{x}^{0}{y}^{4}$

$={x}^{4}+4{x}^{3}y+6{x}^{2}{y}^{2}+4x{y}^{3}+{y}^{4}$ ...(4)

In observing these expressions, you may notice some interesting patterns:

- Power for
*x*starts from n and decreases with every subsequent term until it reaches 0. - Power for
*y*starts from 0 and increases with every subsequent term until it reaches*n*. - The coefficients in the expansion are obtained by using the
*n choose k*

Binomial theorem helps with:

- Solving problems in the field of combinatorics, calculus, algebra, and other areas of mathematics.
- Binomial distribution, which gives the probability of an event returning an outcome of success or failure
- Negative binomial distribution, a probability-based distribution in which the number of events needed to return an outcome of success
- Calculating the coefficient of a given expansion in a binomial multiplication without building a Pascal’s triangle.

## More Examples

**Question 1**

Find the coefficient of *x*^{4} in the expansion of (*x*+5)^{5}

__Solution:__

${\left(x+5\right)}^{5}=\sum _{k=0}^{5}\left(\begin{array}{c}5\\ k\end{array}\right){x}^{5-k}5k$

__Step 1:__ Identify the value of *k *needed for substitution.

${x}^{4}={x}^{5-k}$

Compare the exponents and solve for *k*.

4 = 5-*k*

*k* = 5-4

*k* = 1

The value of *k* is identified as 1.

__Step 2:__

We can then calculate the coefficient as follows:

$\left(\begin{array}{c}5\\ 1\end{array}\right){x}^{5-1}{5}^{1}=\frac{5!}{(5-1)!1!}{x}^{4}{5}^{1}=25{x}^{4}$

**Question 2**

What is the coefficient of *x*^{4} *y*^{15} in the expansion of (*x*+*y*)^{19}

^{${\left(x+y\right)}^{19}=\sum _{k=0}^{19}\left(\begin{array}{c}19\\ k\end{array}\right){x}^{19-k}{y}^{k}$}

__Step 1:__ Identify the value of *k *needed for substitution

${y}^{k}={y}^{15}$

Compare the exponents. We that k = 15, thus, the value of *k* is 15.

__Step 2:__

The coefficient is then calculated as:

$\left(\begin{array}{c}19\\ 15\end{array}\right){x}^{19-15}{y}^{15}=\frac{19!}{(19-15)!15!}{x}^{4}{y}^{15}=3876{x}^{4}{y}^{15}$

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