# 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

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 $\large{n}$ is small. However, can Pascal’s Triangle be used for the expansion of $(1 + b)^{20}$?
• 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{\begin{align} &= 1a^2+2ab+1b^2 \end{align}} \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!

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