 Study S3 A Math Binomial Theorem - Geniebook    # 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 $$(a + b)^2 = a^2 + 2ab + b^2$$. However, for the expansion of $$(a + b)^n$$ 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 $$(a + b)^n$$
• Binomial Theorem

## Expansion Of Algebraic Expressions Of The Form $$(1 + b)^n$$

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

 $$(1 + b)^1$$ \begin{align} &= 1 + b \end{align} $$(1 + b)^2$$ \begin{align} &= 1 + 2b + b^2 \end{align} $$(1 + b)^3$$ \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} $$(1 + b)^4$$ \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.

\begin{align*} (1+b)^0 &= 1\\ (1+b)^1 &= 1+1b\\ (1+b)^2 &= 1+2b+1b^2\\ (1+b)^3 &= 1+3b+3b^2+1b^3\\ (1+b)^4 &= 1+4b+6b^2+4b^3+1b^4\\ (1+b)^5 &= 1+5b+10b^2+10b^3+5b^4+1b^5\\ (1+b)^6 &= 1+6b+15b^2+20b^3+15b^4+6b^5+1b^6 \end{align*}

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

\begin{align*} 1\\ 1 + 1\\ 1 + 2 + 1\\ 1 + 3 + 3 + 1\\ 1 + 4 + 6 + 4 + 1\\ 1 + 5 + 10 + 10 + 5 + 1\\ 1 + 6 + 15 + 20 + 15 + 6 + 1 \end{align*}

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

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 $$n$$ 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 \begin{align} \binom {n}{r} \end{align} 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 \begin{align} \binom {n}{0} \end{align} not \begin{align} \binom {n}{1} \end{align}.
• 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:

\begin{align*} (1+x)^{20} &=\binom{20}{0}+\binom{20}{1}x^1+\binom{20}{2}x^2+\binom{20}{3}x^3 + \cdots \\ &= 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:

\begin{align} (1+y)^{13} &=\binom{13}{0}+\binom{13}{1}y^1+\binom{13}{2}y^2+\binom{13}{3}y^3 + \cdots \\ &= 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 $$(a + b)^n$$?

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

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

Here, $$a^3+3a^{2}b+3ab^2+b^3$$ 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 $$a^3+3a^{2}b+3ab^2+b^3$$.

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

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

## Binomial Theorem From the above investigation, the result of the expansion of $$(a+b)^n$$ is as follows,

\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:

\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 \\ &= 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:

\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 \\ &=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, \begin{align} \binom {n}{r} \end{align}, of a binomial expansion $$(a + b)^n$$

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

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

Keep learning! Keep improving!

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