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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