Step-by-step Guide to Binomial Theorem

Before we start, we need to know that there is such a thing called Pascal’s triangle.

Pascal’s triangle is actually a very fun triangle to play with.

So we start the triangle with one at the top. Then we slide down to both sides and move one in.

For the interior part, we will have to connect the two numbers above together and add them up.

The pattern repeats.

Now, let’s compare the coefficients of the expansions with Pascal’s triangle above.

${{left( a+b right)}^{2}}=%%MATHDISPLAY0%%{{a}^{2}}+%%MATHDISPLAY1%%ab+%%MATHDISPLAY2%%{{b}^{2}}$
${{left( a+b right)}^{3}}=%%MATHDISPLAY3%%{{a}^{3}}+%%MATHDISPLAY4%%{{a}^{2}}b+%%MATHDISPLAY5%%a{{b}^{2}}+%%MATHDISPLAY6%%{{b}^{3}}$
${{left( a+b right)}^{4}}=%%MATHDISPLAY7%%{{a}^{4}}+%%MATHDISPLAY8%%{{a}^{3}}b+%%MATHDISPLAY9%%{{a}^{2}}{{b}^{2}}+%%MATHDISPLAY10%%a{{b}^{3}}+%%MATHDISPLAY11%%{{b}^{4}}$

We could see that there is a similarity in the pattern for the expansion of our expression here. It is pretty obvious how things are working for the coefficients now, but what about the power of the constants? Please note that there is a special pattern of counting for the power of each term in Pascal’s triangle.

${{left( a+b right)}^{4}}=1%%MATHDISPLAY12%%{{b}^{0}}%%MATHDISPLAY13%%{{a}^{3}}%%MATHDISPLAY14%%+6%%MATHDISPLAY15%%{{b}^{2}}%%MATHDISPLAY16%%{{a}^{1}}%%MATHDISPLAY17%%+1%%MATHDISPLAY18%%{{b}^{4}}$

The powers of the term $a$ descends from $4$ to zero while the powers of the other term, $b$ ascends from zero to $4$. 

So, if you want to find, for example, ${{left( a+b right)}^{5}}$, we could use the subsequent level of Pascal’s triangle to help.

To expand ${{left( a+b right)}^{5}}$:

1.

We know that ${{left(a+bright)}^{n}}$ would yield $n+1$ terms. So, we could expect to have six terms after the expansion of ${{left(a+bright)}^{5}}$.
Let's write it as:

${{left( a+b right)}^{5}}=ab+ab+ab+ab+ab+ab$

2.

Then, run down the powers of the terms where the powers of the term $a$ descend from $5$ to zero while the powers of the other term, $b$ would ascend from zero to $5$.

${{left( a+b right)}^{5}}=%%MATHDISPLAY0%%{{b}^{0}}%%MATHDISPLAY1%%{{a}^{4}}%%MATHDISPLAY2%%+%%MATHDISPLAY3%%{{b}^{2}}%%MATHDISPLAY4%%{{a}^{2}}%%MATHDISPLAY5%%+%%MATHDISPLAY6%%{{b}^{4}}%%MATHDISPLAY7%%{{a}^{0}}%%MATHDISPLAY8%%1%%MATHDISPLAY9%%5%%MATHDISPLAY10%%10%%MATHDISPLAY11%%10%%MATHDISPLAY12%%5%%MATHDISPLAY13%%1$${{a}^{0}}{{b}^{5}}$

4.

Simplify the expansion.

${{left( a+b right)}^{5}}={{a}^{5}}+5{{a}^{4}}b+10{{a}^{3}}{{b}^{2}}+10{{a}^{2}}{{b}^{3}}+5a{{b}^{4}}+{{b}^{5}}$

However, it is impossible for us to draw out Pascal’s triangle every time, especially if it involves the expansion of higher powers.

In fact, there is a formula for us to find the binomial coefficients and this formula is called “$n$ choose $r$” , written as ${}^{n}{{c}_{r}}$ or $left( begin{matrix}n \r \end{matrix} right)$.

But how do we identify the terms in Pascal’s triangle? We need to bear in mind that counting always starts from zero.

The $n$ refers to the level, which corresponds to the power of the expansion.

This makes it easy to find the coefficient of a particular term as we can use our calculator to find the value of ${}^{n}{{c}_{r}}$. 

Example: Finding Coefficient of a Specific Term

To find the coefficient of the third term for the expansion ${{left( a+b right)}^{5}}$:

	$begin{aligned}{}^{5}{{C}_{3}}&=left( begin{matrix}5 \3 \end{matrix} right) \ & =frac{5!}{3!left( 5-2 right)!} \& =10 end{aligned}$

For the expansion of ${{(a+b)}^{n}}$ , the general term formula for ${{(r+1)}^{text{th}}}$ term is ${{T}_{r+1}}=left( begin{matrix}n \r \end{matrix} right){{a}^{(n-r)}}{{b}^{r}}$.

Let’s see how we can apply the general term formula in some simple questions.

(1)

Finding the 5th term of the expansion ${{(1+2x)}^{6}}$

$r=4$ (Note: we start counting from zero hence 5th term will have $r=4$)
5th term, ${{T}_{5}}={{T}_{4+1}}=left( begin{matrix}
6 \
4 \
end{matrix} right){{1}^{(6-4)}}{{left( 2x right)}^{4}}=240{{x}^{3}}$

(2)

Finding the 7th term of the expansion ${{(3+x)}^{8}}$

$r=6$ , 7th term, ${{T}_{7}}={{T}_{6+1}}=left( begin{matrix}
8 \
6 \
end{matrix} right){{3}^{(8-6)}}{{left( x right)}^{6}}=504{{x}^{3}}$

(3)

Finding the 5th term of the expansion ${{(2x+1)}^{6}}$

$r=4$, 5th term, ${{T}_{5}}={{T}_{4+1}}=left( begin{matrix}
6 \
4 \
end{matrix} right){{left( 2x right)}^{(6-4)}}{{left( 1 right)}^{4}}=60{{x}^{2}}$

Note that the 5th term of (1) and (3) are different. Why is it different when ${{(1+2x)}^{6}}={{(2x+1)}^{6}}$? It’s because the power of expansion is different. For (1) we ascend the power of $x$ as for (3) we descend the power of $x$. This shows that, though the general term formula is convenient, it’s not all that useful if the context is not given.

This is a brief overview of Pascal’s triangle and Binomial Theorem. Download free printable worksheet here for some extra practice. Now, let’s see how we can solve problems involving binomial theorem in the subsequent questions.