# Binomial Expansions

The sum $a+b$ is called a binomial as it contains two terms.
Any expression of the form $(a+b)^n$ is called a power of a binomial.

All binomials raised to power can be expanded using the same general principles. In this lesson, therefore, we consider the expansion of the general expression $(a+b)^n$ where $n \in \mathbb{N}$.

Consider the following algebraic expressions.
\begin{align} \displaystyle (a+b)^1 &= a^1 + b^1 \\ &= a+b \\ (a+b)^2 &= a^2 + 2a^1b^1 + b^2 \\ &= a^2 + 2ab + b^2 \\ (a+b)^3 &= a^3 + 3a^2b^1 + 3a^1b^2 + b^3 \\ &= a^3 + 3a^2b + 3ab^2 + b^3 \\ \end{align}

In the theory of Pascal’s Triangle, the coefficients of binomial expansions come from Pascal’s Triangle.
\begin{matrix}
n=0&&&&&&1 \\
n=1&&&&&1&&1 \\
n=2&&&&1&&2&&1 \\
n=3&&&1&&3&&3&&1 \\
n=4&&1&&4&&6&&4&&1 \\
\end{matrix}
Let’s consider the expansion of $(a+b)^4$.
This expansion produces $5$ terms; $a^4;a^3b^1;a^2b^2;a^1b^3;b^4$.
Their corresponding coefficients are $1;4;6;4;1$, respectively as per Pascal’s Triangle.
Therefore we can obtain the expansion of $(a+b)^4$ by combining these terms and coefficients.
\begin{align} \displaystyle (a+b)^4 &= 1a^4 + 4a^3b^1 + 6a^2b^2 + 4a^1b^3 + 1b^4 \\ &= a^4 + 4a^3b + 6a^2b^2 + 4ab^3 + b^4 \\ \end{align}

### Example 1

Expand $(a+b)^5$.

\begin{align} \displaystyle \begin{matrix} n=0&&&&&&1 \\ n=1&&&&&1&&1 \\ n=2&&&&1&&2&&1 \\ n=3&&&1&&3&&3&&1 \\ n=4&&1&&4&&6&&4&&1 \\ n=5&1&&5&&10&&10&&5&&1 \\ \end{matrix} \end{align}

\begin{align} \displaystyle (a+b)^5 &= 1a^5 + 5a^4b^1 + 10a^3b^2 + 10a^2b^3 + 5a^1b^4 + 1b^5 \\ &= a^5 + 5a^4b + 10a^3b^2 + 10a^2b^3 + 5ab^4 + b^5 \end{align}

### Example 2

Expand $(a-b)^5$.

\begin{align} \displaystyle \begin{matrix} n=0&&&&&&1 \\ n=1&&&&&1&&1 \\ n=2&&&&1&&2&&1 \\ n=3&&&1&&3&&3&&1 \\ n=4&&1&&4&&6&&4&&1 \\ n=5&1&&5&&10&&10&&5&&1 \\ \end{matrix} \end{align}

\begin{align} \displaystyle (a-b)^5 &= 1a^5 + 5a^4(-b)^1 + 10a^3(-b)^2 + 10a^2(-b)^3 + 5a^1(-b)^4 + 1(-b)^5 \\ &= a^5-5a^4b + 10a^3b^2-10a^2b^3 + 5ab^4-b^5 \end{align}

### Example 3

Expand $(a+2b)^5$.

\begin{align} \displaystyle \begin{matrix} n=0&&&&&&1 \\ n=1&&&&&1&&1 \\ n=2&&&&1&&2&&1 \\ n=3&&&1&&3&&3&&1 \\ n=4&&1&&4&&6&&4&&1 \\ n=5&1&&5&&10&&10&&5&&1 \\ \end{matrix} \end{align}

\begin{align} \displaystyle (a+2b)^5 &= 1a^5 + 5a^4(2b)^1 + 10a^3(2b)^2 + 10a^2(2b)^3 + 5a^1(2b)^4 + 1(2b)^5 \\ &= a^5 + 5a^4 \times 2b + 10a^3 \times 4b^2 + 10a^2 \times 8b^3 + 5a \times 16b^4 + 32b^5 \\ &= a^5 + 10a^4b + 40a^3b^2 + 80a^2b^3 + 80ab^4 + 32b^5 \end{align}

### Example 4

Expand $\bigg(a+\dfrac{1}{a}\bigg)^3$.

\begin{align} \displaystyle \begin{matrix} n=0&&&&&1 \\ n=1&&&&1&&1 \\ n=2&&&1&&2&&1 \\ n=3&&1&&3&&3&&1 \\ \end{matrix} \end{align}

\begin{align} \displaystyle \bigg(a+\dfrac{1}{a}\bigg)^3 &= 1a^3 + 3a^2 \times \dfrac{1}{a^1} + 3a^1 \times \dfrac{1}{a^2} + \dfrac{1}{a^3} \\ &= a^3 + 3a + \dfrac{3}{a} + \dfrac{1}{a^3} \end{align}

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