# Binomial Theorem

In this article, we will learn about Binomial Theorem. So far, we have learnt how to expand the square of a binomial using algebraic identities like \(\displaystyle{(a + b)^2 = a^2 + 2ab + b^2}\). However, for the expansion of \(\displaystyle{(a + b)^n}\) where *\(n\)* is a positive integer greater than \(2\), how do we get the result of the expansion?

We will be touching on the following concepts:

- Pascal’s Triangle
- Binomial Coefficients
- Binomial Expansion of \(\displaystyle{(a + b)^n}\)
- Binomial Theorem

**Expansion of Algebraic Expressions of the Form (1 + b)**^{n}

^{n}

The table below shows the expansions of algebraic expressions of the form \(\small{\displaystyle{(1 + b)^n}}\), where \(\small{n = 1, 2, 3, 4}\).

\(\small{\displaystyle{(1 + b)^1}}\) | \(\small{\begin{align}&=\displaystyle{1 + b}\end{align}}\) |

\(\small{\displaystyle{(1 + b)^2}}\) | \(\small{\begin{align}&=\displaystyle{ 1 + 2b + b^2}\end{align}}\) |

\(\small{\displaystyle{(1 + b)^3}}\) | \(\small{\begin{align} &= (1 + b)(1 + b)^2 \\ &= (1 + b)(1 + 2b + b^2) \\ &= 1 + 2b + b^2 + b + 2b^2+ b^3 \\ &= 1 + 3b + 3b^2 + b^3 \\ \end{align}}\) |

\(\small{\displaystyle{(1 + b)^4}}\) | \(\small{\begin{align} &= (1 + b)(1 + 3b + 3b^2 + b^3) \\ &= 1 + 3b + 3b^2 + b^3 + b + 3b^2 + 3b^3 + b^4 \\ &= 1 + 4b + 6b^2 + 4b^3 + b^4 \end{align}}\) |

## Investigation: Introduction to Pascal’s Triangle

Pascal was a French mathematician and scientist who formulated what became known as Pascal’s Principle of Pressure. The SI unit for pressure is pascal (\(Pa\)). Pascal is also famously known for Pascal’s Triangle which is used to derive binomial coefficients in binomial expansions.

\(\small{\begin{matrix} & (1+b)^0 = 1 &\\[2ex] & (1+b)^1 = 1+1b &\\[2ex] & (1+b)^2 = 1+2b+1b^2 &\\[2ex] & (1+b)^3 = 1+3b+3b^2+1b^3 &\\[2ex] & (1+b)^4 = 1+4b+6b^2+4b^3+1b^4 &\\[2ex] & (1+b)^5 = 1+5b+10b^2+10b^3+5b^4+1b^5 &\\[2ex] & (1+b)^6 = 1+6b+15b^2+20b^3+15b^4+6b^5+1b^6 & \end{matrix}}\)

The triangle above shows the expansions of algebraic expressions of the form \(\small{(1 + b)^n}\), where \(\small{n = 1, 2, 3, 4, 5, 6}\). If we reduce the triangle to just the coefficients of each term, we will get Pascal’s Triangle, as seen below:

\(\small{\begin{matrix} 1\\[2ex] 1 + 1\\[2ex] 1 + 2 + 1\\[2ex] 1 + 3 + 3 + 1\\[2ex] 1 + 4 + 6 + 4 + 1\\[2ex] 1 + 5 + 10 + 10 + 5 + 1\\[2ex] 1 + 6 + 15 + 20 + 15 + 6 + 1 \end{matrix}}\)

Notice that each new entry, *\(\large{x}\)*, of each subsequent row forms a triangle with the number above and to the left, *\(\large{a_{_L}}\)*, and the number above and to the right, *\(\large{a_{_R}}\)*. We obtain *\(\large{x}\)* by taking the sum of *\(\large{a_{_L}}\)* and *\(\large{a_{_R}}\)*.

For example, in the fifth row, *\(6\)* is obtained by taking the sum of *\(3\)* and *\(3\)* from the fourth row.

- From this investigation, we observe that Pascal’s Triangle can be used to expand \((1 + b)^n\) if
*\(\large{n}\)*isHowever, can Pascal’s Triangle be used for the expansion of \((1 + b)^{20}\)?__small__.**No.**Pascal's Triangle is not possible for the expansion of \((1 + b)^{20}\). This is because the**expansion will be extremely tedious**and we would need to start from \((1 + b)^{1}\).

Hence, we need to explore a different method to obtain binomial expansions with greater powers.

## Introduction to Binomial Coefficient** **

- The notation \(\large{\binom {n}{r}}\) read as “\(n \;\text{choose} \;r\)”, represents a binomial coefficient in the expansion of \((1 + b)^n\).

*\(n\)* represents the power and the *\(r\)* represents the position, starting from \(0\).

- The first coefficient is \(\large{\binom {n}{0}}\) not \(\large{\binom {n}{1}}\).

- In general the binomial expansion of \((1 + b)^n\) is

\(\begin{align*} (1+b)^n=\binom{n}{0}+\binom{n}{1}b+\binom{n}{2}b^2+...+\binom{n}{n}b^n \end{align*}\)

As you can see in the above equation, the first position of the binomial formula is \(0\), followed by \(1, 2, \cdots n\). Hence, the binomial coefficients are \(\begin{align*} \binom{n}{0},\;\binom{n}{1}, \;\cdots \;,\binom{n}{n} \end{align*}\).

**Question 1:**

Find, in ascending powers of *\(x\)*, the first four terms in the expansion of \((1+x)^{20}\).

**Solution:**

\(\small{\begin{align*} (1+x)^{20} &=\binom{20}{0}+\binom{20}{1}x^1+\binom{20}{2}x^2+\binom{20}{3}x^3 + \cdots \\[2ex] &= 1+20x+190x^2+1140x^3+\cdots \end{align*}}\)

**Question 2:**

Find, in ascending powers of \(y\), the first four terms in the expansion of \((1+y)^{13}\).

**Solution:**

\(\small{\begin{align} (1+y)^{13} &=\binom{13}{0}+\binom{13}{1}y^1+\binom{13}{2}y^2+\binom{13}{3}y^3 + \cdots \\[2ex] &= 1+13y+78y^2+286y^3+\cdots \end{align}}\)

### Binomial Expansion of (a + b)^{n}

We have learnt how to expand \((1 + b)^n\).

What about \((a + b)^n\)?

\(\small{(a+b)^1}\) | \(\small{\begin{align} &= 1a+1b \end{align}}\) |
---|---|

\(\small{(a+b)^2}\) | \(\small{\begin{align} &= 1a^2+2ab+1b^2 \end{align}}\) |

\(\small{(a+b)^3}\) | \(\small{\begin{align} &=1a^3+3a^{2}b+3ab^2+1b^3 \end{align}}\) |

From the table, \(\small{\begin{align} (a+b)^3 &=a^3+3a^{2}b+3ab^2+b^3 \end{align}}\).

Here, \(\small{a^3+3a^{2}b+3ab^2+b^3 }\) has the binomial coefficients \(1, \;3, \;3, \;1,\) which follows the fourth row of Pascal’s Triangle. Notice that the power of \(a\) decreases and power of \(b\) increases in \(\small{a^3+3a^{2}b+3ab^2+b^3 }\).

Hence, we can deduce that the binomial expansion of \((a+b)^4\) is

\(\small{(a+b)^4= 1a^4+4a^{3}b+6a^{2}b^{2}+4ab^3+b^4}\).

## Binomial Theorem

From the above investigation, the result of the expansion of \((a+b)^n\) is as follows,

\(\small{\begin{align} (a+b)^n &= \binom{n}{0}a+\binom{n}{1}a^{(n-1)}b^1+\binom{n}{2}a^{(n-2)}b^2+\cdots+\binom{n}{r}a^{(n-r)}b^r+\cdots+\binom{n}{n}b^n \\ \end{align}}\)

**Question 3:**

Find, in ascending powers of \(x\), the first four terms in the expansion of \((3+2x)^7\).

**Solution:**

\(\small{\begin{align} (3+2x)^7&=\binom{7}{0}(3)^7(2x)^0 +\binom{7}{1}(3)^{6}(2x)^1+\binom{7}{2}(3)^{5}(2x)^2+\binom{7}{3}(3)^{4}(2x)^3+\cdots \\[2ex] &= 2187+10206x+20412x^2+22680x^3+... \end{align}}\)

**Question 4:**

Find, in ascending powers of \(y\), the first four terms in the expansion of \((2+3y)^5\).

**Solution:**

\(\small{\begin{align} (2+3y)^5 &= \binom{5}{0}(2)^{5}(3y)^0+\binom{5}{1}(2)^{4}(3y)^1+\binom{5}{2}(2)^{3}(3y)^2+\binom{5}{3}(2)^{2}(3y)^3+\cdots \\[2ex] &=32+240y+720y^2+1080y^3+\cdots \end{align}}\)

## Conclusion

In this article, we have learnt about Pascal’s Triangle and how it relates to the binomial coefficients, \( \binom {n}{r}\), of a binomial expansion \((a + b)^n\).

An important reminder when performing binomial expansion is that the first binomial coefficient is always \(\binom {n}{0}\), while the last binomial coefficient is always \(\binom {n}{n}\).

Also make sure that you are careful in your manipulation when simplifying the expression!

Keep learning! Keep improving!