The following page will help you with questions about the Binomial Theorem (or Binomial Expansion). The Binomial Theorem is used for expanding brackets in the form (a + b)n
On this page, you should learn to
- Expand \((a+b)^n\) using Pascal's triangle and \(^nC_r\)
The following videos will help you understand all the concepts from this page
Spotting the Pattern
At first look, the formula for the Binomial Theorem looks quite complicated
\(\left(a+b\right)^{n}=a^{n}+\left(\begin{array}{c} n\\ 1 \end{array}\right) a^{n-1}b+...+\left(\begin{array}{c} n\\ r \end{array}\right)a^{n-r}b^{r}+...+b^{n} \)
The formula is in the information booklet, so no need to try and lean it by heart! Applying it is not too difficult, but before we do that, it is important that we understand the pattern (or patterns) that lie behind it.
The following video shows you the patterns behind the Binomial Theorem
Using Pascal's Triangle
The numbers in Pascal's triangle might well be familiar to you. Each number can be calculated by adding to the numbers immediately above it. As we saw in the previous video, the numbers in Pascal's triangle make up part of the pattern in the expansion of \((a + b)^n\)
We can use Pascal's triangle to help us expand brackets in the form \((a + b)^n\) . Watch the following video to see how to expand \((1 +4x)^3\).
What are Combinations?
The numbers that appear in Pascal's triangle can also be found using combinations. HL students should know that a combination is a way of selecting items from a certain number of objects. We can write combinations in different ways \(_{n} P_{r} \) or \(\binom {n} {r}\)
The formula for combinations can be found in the information booklet \(\binom {n} {r} = \frac{n!}{r!(n-r)!}\)
The diagram below shows how each number in Pascal's triangle can be written as a combination.
Here's a video to explain in detail how we can find each number in Pascal's triangle with a combination.
Print from here
Here is a short quiz to test your understanding of combinations
START QUIZ!
The 4th row of Pascal's triangle is
1 4 a 4 1
What is the value of a?
a =
The 5th row of Pascal's triangle is
1 5 a a 5 1
What is the value of a?
a =
Evaluate \(\large^6C_2\)
\(\large^6C_2\) =
\(\large^nC_r=\frac{n!}{(n-r)!\ r!}\)
\(\large^6C_2=\frac{6!}{4!\ 2!}=\frac{6\cdot 5}{2}=15\)
Evaluate \(\large^8C_6\)
\(\large^8C_6\) =
\(\large^nC_r=\frac{n!}{(n-r)!\ r!}\)
\(\large^8C_6=\frac{8!}{2!\ 6!}=\frac{8\cdot 7}{2}=28\)
Evaluate \(\large^{10}C_3\)
\(\large^{10}C_3\) =
\(\large^nC_r=\frac{n!}{(n-r)!\ r!}\)
\(\large^{10}C_3=\frac{10!}{7!\ 3!}=\frac{10\cdot 9 \cdot 8}{3\cdot 2}=120\)
Evaluate \(\large^8C_8\)
\(\large^8C_8\) =
\(\large^nC_r=\frac{n!}{(n-r)!\ r!}\)
\(\large^8C_8=\frac{8!}{0!\ 8!}=1\)
Evaluate \(\large^7C_6\)
\(\large^7C_6\) =
\(\large^nC_r=\frac{n!}{(n-r)!\ r!}\)
\(\large^7C_6=\frac{7!}{1!\ 6!}=7\)
Given that r>4, evaluate r , if \(\large^6C_r=20\)
r =
\(\large^nC_r=\frac{n!}{(n-r)!\ r!}\)
You can use a calculator to work this out using trial and error, or the table function
\(\large^6C_3=\frac{6!}{3!\ 3!}=\frac{6\cdot 5\cdot 4}{3\cdot 2}=20\)
Evaluate r , if \(\large^8C_r=56\)
r =
\(\large^nC_r=\frac{n!}{(n-r)!\ r!}\)
You can use a calculator to work this out using trial and error, or the table function
\(\large^8C_5=\frac{8!}{3!\ 5!}=\frac{8\cdot 7\cdot 6}{3\cdot 2}=56\)
Evaluate n , if \(\large^nC_5=21\)
r =
\(\large^nC_r=\frac{n!}{(n-r)!\ r!}\)
You can use a calculator to work this out using trial and error, or the table function
\(\large^7C_5=\frac{7!}{2!\ 5!}=\frac{7\cdot 6}{2}=21\)
Here is a quiz that practises the Binomial expansion (a + b)n
START QUIZ!
4 can be written as a combination \(\left(\begin{array}{c} n\\ r \end{array}\right)\)What are n and r?
\(\left(\begin{array}{c} 4\\ 1 \end{array}\right)\)or \(\left(\begin{array}{c} 4\\ 3 \end{array}\right)\)
21 can be written as a combination \(\left(\begin{array}{c} n\\ r \end{array}\right)\)What are n and r?
n= r =
\(\left(\begin{array}{c} 7\\ 2 \end{array}\right)\)or \(\left(\begin{array}{c} 7\\ 5 \end{array}\right)\)
\(\left(\begin{array}{c} 8\\ 2 \end{array}\right)\)is equal to another combination \(\left(\begin{array}{c} n\\ r \end{array}\right)\). What are n and r?
n= r=
Look at the symmetry of Pascal's triangle
Put these values in order of size from smallest to largest
\(\left(\begin{array}{c} 10\\ 8 \end{array}\right), \left(\begin{array}{c} 7\\ 5 \end{array}\right), \left(\begin{array}{c} 6\\ 2 \end{array}\right), \left(\begin{array}{c} 8\\ 2 \end{array}\right), \left(\begin{array}{c} 6\\ 3 \end{array}\right)\)
\(\left(\begin{array}{c} 6\\ 2\end{array}\right)=15, \left(\begin{array}{c} 6\\ 3 \end{array}\right)=20, \left(\begin{array}{c} 7\\ 5 \end{array}\right)=21, \left(\begin{array}{c} 8\\ 2 \end{array}\right)=28, \left(\begin{array}{c} 10\\ 8 \end{array}\right)=45\)
The \({x^2}\) term in the expansion \({(1 - 2x)^6}\) is \({\textbf{a}x^2}\) . Write down the value of a
a =
\(\begin{pmatrix} 6 \\ 2 \end{pmatrix}(-2x)^21^4=15\times4x^2\times1\)
The expansion of \({(2 + 3x)^3}\)is \({\textbf{a}x^3+54x^2+\textbf{b}x+8}\) . Find a and b
a =
b =\(x^3\: term = \begin{pmatrix} 3 \\ 3 \end{pmatrix}(2)^0(3x)^3=1\times1\times(3x)^3 = 27x^3 \)
\(x \:term= \begin{pmatrix} 3 \\ 1 \end{pmatrix}(2)^2(3x)^1=3\times4\times(3x)^1 = 36x\)
The expansion of \(\) \( (2x-{{ 1}\over x})^5\) is \(32x^5 -80x^3+\textbf{a}x +{ \textbf{b}\over x}+ {10 \over x^3}-{1 \over x^5}\). Find a and b.
a =
b =\(x\: term = \begin{pmatrix} 5 \\ 3 \end{pmatrix}(2x)^3({-1 \over x})^2=10\times8x^3\times{1 \over x^2} = 80x \)
\({1 \over x}\: term = \begin{pmatrix} 5 \\ 2 \end{pmatrix}(2x)^2({-1 \over x})^3=10\times4x^2\times{-1 \over x^3} = -40x \)
Find the \(x^2\) term in the expansion of \(\) \( (x^2+{{ 3}\over x})^4\)is \(\textbf{a}x^2\). Find a
To find the \(x^2\) term you need to think about what powers you need to raise \( x^2\) and \( {3\over x}\)
Try the different combinations
- 4 and 0
- 3 and 1
- 2 and 2
- 1 and 3
- 0 and 4
\(x^2\: term = \begin{pmatrix} 4 \\ 2 \end{pmatrix}(x^2)^2({3 \over x})^2=6\times x^4\times{9 \over x^2} = 54x^2 \)
\((1 + x)(2 + x^2)^3 = \textbf{a} + 8x + \textbf{b}x^2 + ...\) Find a and b
a = , b =
\((1 + x)(2 + x^2)^3 =(1 + x)(8 + 12x^2 + 6x^4+x^6)\)
Find the term independent of x in the expansion \((x+ {2\over x})^6\)
You can try to expand each term out, but you can save time by thinking about when the x term and \({1 \over x}\)will cancel out. \(\left(\! \begin{array}{c} 6 \\ 3 \end{array} \!\right) x^3(\frac{2}{x})^3\)
The values of the third row of Pascal's triangle are given below
a) Write down the values in the fourth row of Pascal's triangle
b) Hence or otherwise, find the term in x² in the expansion of \((3x+2)^4\)
Hint
Full Solution
a) Expand (x - 3)4 and simplify your result
b) Hence find the x3 term in (x + 2)(x - 3)4 .
Hint
Full Solution
Find the term independent of x in the expansion \((2x- \frac{3}{x^2})^6\)
Hint
Full Solution
How much of Binomial Theorem SL have you understood?
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