Tim Gan Math
O Level (Sec 3 & 4)Additional Mathematics

Binomial Theorem

Master the Binomial Theorem with this comprehensive study guide. Learn Pascal's Triangle, binomial coefficients, and how to expand binomial expressions. Includes 8 worked examples with step-by-step video solutions.

Timothy Gan
May 19, 2021
Download Free Worksheet (PDF)

Understanding the Binomial Theorem

Hi students, let's explore the concept of the Binomial Theorem today! From my experience, this is usually one of the most confusing topics as students approach it with rote memorization instead of seeing the pattern in the expansions.
In this article, I will explain how the formula comes about and provide some examples for you to practice.
Binomial Theorem is a very common topic that can be found in many mathematics syllabi such as the IGCSE Additional Mathematics (0606) and Singapore SEAB Additional Mathematics.

Study Guide - Understanding Binomial Theorem

Understanding Binomial Expansions

From previous Elementary Mathematics' syllabus, we have already learnt the expansion of $(a+b)^2$.
$(a+b)^2 = a^2 + 2ab + b^2$
To find $(a+b)^3$, we will need to multiply $(a+b)^2$ with $(a+b)$ and expand the products:
$(a+b)^3 = (a+b)^2(a+b) = (a^2 + 2ab + b^2)(a+b) = a^3 + 3a^2b + 3ab^2 + b^3$
Similarly for $(a+b)^4$:
$(a+b)^4 = a^4 + 4a^3b + 6a^2b^2 + 4ab^3 + b^4$
Did you notice that there is some form of pattern going on with the coefficients? This is the essence of the binomial expansion.

Pascal's Triangle

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. We start the triangle with 1 at the top. Then we slide down to both sides and move 1 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:
$(a+b)^2 = 1a^2 + 2ab + 1b^2$ $(a+b)^3 = 1a^3 + 3a^2b + 3ab^2 + 1b^3$ $(a+b)^4 = 1a^4 + 4a^3b + 6a^2b^2 + 4ab^3 + 1b^4$
We could see that there is a similarity in the pattern for the expansion. The powers of the term 'a' descends from 4 to zero while the powers of the other term 'b' ascends from zero to 4.
Pascal's Triangle diagram showing the pattern of coefficients

How to Use Pascal's Triangle

If you want to find, for example, $(a+b)^5$, we could use the subsequent level of Pascal's triangle to help.
To expand $(a+b)^5$:
1. We know that $(a+b)^n$ would yield $n+1$ terms. So, we could expect to have six terms after the expansion of $(a+b)^5$.
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.
$(a+b)^5 = a^5b^0 + a^4b^1 + a^3b^2 + a^2b^3 + a^1b^4 + a^0b^5$
3. Fill in the coefficients using Pascal's triangle:
$(a+b)^5 = 1a^5 + 5a^4b + 10a^3b^2 + 10a^2b^3 + 5ab^4 + b^5$
However, it is impossible for us to draw out Pascal's triangle every time, especially if it involves the expansion of higher powers.

Binomial Coefficients

In fact, there is a formula for us to find the binomial coefficients. This formula is called 'n choose r', written as $^nC_r$ or $\binom{n}{r}$.
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 $^nC_r$.
For the expansion of $(a+b)^n$, the general term formula for the $(r+1)$th term is:
$T_{r+1} = \binom{n}{r} \times a^{(n-r)} \times b^r$
Important: Counting always starts from zero!

Practice Questions with Video Solutions

Watch step-by-step video explanations for each question to master the concepts

Question 1Video Solution
Binomial Expansion involving Linear Equation

Question

Find, in ascending powers of $x$, the first four terms in the expansion of $(2-3x)^6$.

Video Solution

Step-by-Step Solution

  1. 1From the question, we know that $a=2$, $b=-3x$ and $n=6$.
  2. 2Substitute the values into the formula and expand it.
  3. 3$(2-3x)^6 = \binom{6}{0}(2)^6(-3x)^0 + \binom{6}{1}(2)^5(-3x)^1 + \binom{6}{2}(2)^4(-3x)^2 + \binom{6}{3}(2)^3(-3x)^3 + \ldots$
  4. 4$= (1)(2)^6(1) + 6(2)^5(-3x) + 15(2)^4(-3)^2x^2 + 20(2)^3(-3)^3x^3 + \ldots$
  5. 5$= 64 - 576x + 2160x^2 - 4320x^3 + \ldots$
Answer:$64 - 576x + 2160x^2 - 4320x^3 + \ldots$
Question 2Video Solution
Binomial Expansion involving Fraction

Question

Find, in ascending powers of $x$, the first four terms in the expansion of $\left(2x + \frac{y}{4}\right)^6$.

Video Solution

Step-by-Step Solution

  1. 1From the question, we know that $a=2x$, $b=\frac{y}{4}$ and $n=6$.
  2. 2Substitute the values into the formula and expand it.
  3. 3$\left(2x + \frac{y}{4}\right)^6 = \binom{6}{0}(2x)^0\left(\frac{y}{4}\right)^6 + \binom{6}{1}(2x)^1\left(\frac{y}{4}\right)^5 + \binom{6}{2}(2x)^2\left(\frac{y}{4}\right)^4 + \binom{6}{3}(2x)^3\left(\frac{y}{4}\right)^3 + \ldots$
  4. 4$= (1)(1)\frac{y^6}{4^6} + \frac{6(2)}{4^5}xy^5 + \frac{15(2)^2}{4^4}x^2y^4 + \frac{20(2)^3}{4^3}x^3y^3 + \ldots$
  5. 5$= \frac{1}{4096}y^6 + \frac{3}{256}xy^5 + \frac{15}{64}x^2y^4 + \frac{5}{2}x^3y^3 + \ldots$
Answer:$\frac{1}{4096}y^6 + \frac{3}{256}xy^5 + \frac{15}{64}x^2y^4 + \frac{5}{2}x^3y^3 + \ldots$
Question 3Video Solution
Evaluating Binomial Coefficients Without Calculator

Question

Evaluate the following without using a calculator: (a) $\binom{10}{3}$ (b) $\binom{n+2}{2}$ in terms of $n$ (c) $\binom{n+1}{n}$ in terms of $n$ (d) $\binom{n}{n}$ in terms of $n$

Video Solution

Step-by-Step Solution

  1. 1(a) $\binom{10}{3} = \frac{10!}{3!(10-3)!} = \frac{10!}{3!7!} = \frac{10 \cdot 9 \cdot 8 \cdot 7!}{3 \cdot 2 \cdot 1 \cdot 7!} = 120$
  2. 2(b) $\binom{n+2}{2} = \frac{(n+2)!}{2!(n+2-2)!} = \frac{(n+2)!}{2 \cdot n!} = \frac{(n+2)(n+1)n!}{2 \cdot n!} = \frac{(n+2)(n+1)}{2}$
  3. 3(c) $\binom{n+1}{n} = \frac{(n+1)!}{n!(n+1-n)!} = \frac{(n+1)!}{n!1!} = \frac{(n+1)n!}{n!} = n+1$
  4. 4(d) $\binom{n}{n} = \frac{n!}{n!(n-n)!} = \frac{n!}{n!0!} = 1$
Answer:(a) $120$, (b) $\frac{(n+2)(n+1)}{2}$, (c) $n+1$, (d) $1$
Question 4Video Solution
Finding the Coefficient of a Specific Term

Question

Find, in ascending powers of $x$, the first three terms in the expansion of $\left(2 - \frac{x}{2}\right)^6$. Hence, find the coefficient of $x^2$ in the expansion of $(x+1)^2\left(2 - \frac{x}{2}\right)^6$.

Video Solution

Step-by-Step Solution

  1. 1First, find the first three terms of $\left(2 - \frac{x}{2}\right)^6$:
  2. 2$\left(2 - \frac{x}{2}\right)^6 = \binom{6}{0}(2)^6\left(-\frac{x}{2}\right)^0 + \binom{6}{1}(2)^5\left(-\frac{x}{2}\right)^1 + \binom{6}{2}(2)^4\left(-\frac{x}{2}\right)^2 + \ldots$
  3. 3$= 2^6 + 6(2)^5\left(-\frac{1}{2}\right)x + 15(2)^4\left(-\frac{1}{2}\right)^2x^2 + \ldots$
  4. 4$= 64 - 96x + 60x^2 + \ldots$
  5. 5Then, multiply $(x+1)^2$ to find the coefficient of $x^2$:
  6. 6$(x+1)^2\left(2 - \frac{x}{2}\right)^6 = (x^2 + 2x + 1)(64 - 96x + 60x^2 + \ldots)$
  7. 7We only need terms that produce $x^2$: $64x^2 - 96x(2x) + 1(60x^2)$
  8. 8$= 64x^2 - 192x^2 + 60x^2 = -68x^2$
  9. 9Therefore, coefficient of $x^2 = -68$
Answer:First three terms: $64 - 96x + 60x^2 + \ldots$; Coefficient of $x^2$ in final expansion: $-68$
Question 5Video Solution
Finding Unknown Powers

Question

In the expansion of $(1-2x)^n$, where $n$ is a positive integer, in the ascending powers of $x$, the coefficient of the third term is $112$. Find the value of $n$.

Video Solution

Step-by-Step Solution

  1. 1Expand the term $(1-2x)^n$ to find the third term:
  2. 2$(1-2x)^n = \binom{n}{0}(1)^n(-2x)^0 + \binom{n}{1}(1)^{n-1}(-2x)^1 + \binom{n}{2}(1)^{n-2}(-2x)^2 + \ldots$
  3. 3Third term $= \binom{n}{2}(-2x)^2 = \left[\frac{n(n-1)}{2!}\right](-2)^2x^2$
  4. 4Compare the coefficient of the third term with $112$:
  5. 5$\left[\frac{n(n-1)}{2}\right] \cdot 4 = 112$
  6. 6$2n(n-1) = 112$
  7. 7$n(n-1) = 56$
  8. 8$n^2 - n - 56 = 0$
  9. 9$(n-8)(n+7) = 0$
  10. 10Therefore $n = 8$ ($n = -7$ is rejected as $n > 0$)
Answer:$n = 8$
Question 6Video Solution
Finding a Term Independent of x

Question

Find the term independent of $x$ in the expansion of $\left(x^3 - \frac{1}{2x}\right)^{12}$.

Video Solution

Step-by-Step Solution

  1. 1Note that term independent of $x$ refers to the term where the degree of $x = 0$, i.e., $x^0$.
  2. 2Find the general term of $\left(x^3 - \frac{1}{2x}\right)^{12}$:
  3. 3$\left(x^3 - \frac{1}{2x}\right)^{12} = \ldots + \binom{12}{r}(x^3)^{12-r}\left(-\frac{1}{2x}\right)^r + \ldots$
  4. 4$= \binom{12}{r}x^{3(12-r)}\left(-\frac{1}{2}\right)^r(x^{-1})^r$
  5. 5$= \binom{12}{r}\left(-\frac{1}{2}\right)^rx^{36-3r-r}$
  6. 6$= \binom{12}{r}\left(-\frac{1}{2}\right)^rx^{36-4r}$
  7. 7For the power of $x$ to be zero: $x^0 = x^{36-4r}$
  8. 8$0 = 36 - 4r$
  9. 9$r = 9$
  10. 10Substitute $r=9$ into the general term:
  11. 11$\binom{12}{9}\left(-\frac{1}{2}\right)^9 = -\frac{55}{128}$
Answer:$-\frac{55}{128}$
Question 7Video Solution
Finding Value Given Ratio of Coefficients

Question

Find the value of $n$ in $(1+2x)^n$, given that the coefficients of $x^2$ and $x^3$ are in the ratio of $3:14$.

Video Solution

Step-by-Step Solution

  1. 1Write down the general form of $(1+2x)^n$:
  2. 2$(1+2x)^n = \binom{n}{r}(1)^{n-r}(2x)^r$
  3. 3General form $= \binom{n}{r}2^rx^r$
  4. 4Find the coefficient of $x^2$; Substitute $r=2$:
  5. 5Coefficient of $x^2 = \binom{n}{2}2^2 = \left[\frac{n(n-1)}{2!}\right](4) = 2n(n-1)$
  6. 6Find the coefficient of $x^3$; Substitute $r=3$:
  7. 7Coefficient of $x^3 = \binom{n}{3}2^3 = \left[\frac{n(n-1)(n-2)}{3!}\right](8) = \frac{4}{3}n(n-1)(n-2)$
  8. 8Given that coefficient of $x^2 : x^3 = 3:14$
  9. 9$\frac{2n(n-1)}{\frac{4}{3}n(n-1)(n-2)} = \frac{3}{14}$
  10. 10$\frac{2}{\frac{4}{3}(n-2)} = \frac{3}{14}$
  11. 11$2(14) = 3\left[\frac{4(n-2)}{3}\right]$
  12. 12$28 = 4(n-2)$
  13. 13$n = \frac{28}{4} + 2 = 9$
Answer:$n = 9$
Question 8Video Solution
Application of Binomial Theorem

Question

Obtain the first 4 terms in the expansion, in ascending powers of $x$, of $(1-4x)^5$. Hence, find the value of $(0.96)^5$ correct to 3 decimal places.

Video Solution

Step-by-Step Solution

  1. 1Expand the term $(1-4x)^5$ to find the first four terms:
  2. 2$(1-4x)^5 = \binom{5}{0}(1)^5(-4x)^0 + \binom{5}{1}(1)^4(-4x)^1 + \binom{5}{2}(1)^3(-4x)^2 + \binom{5}{3}(1)^2(-4x)^3 + \ldots$
  3. 3$= 1 \cdot 1 \cdot 1 + 5 \cdot 1 \cdot (-4x) + 10(1)(16x^2) + 10(1)(-64)x^3 + \ldots$
  4. 4$= 1 - 20x + 160x^2 - 640x^3 + \ldots$
  5. 5Compare $(1-4x)^5$ with $(0.96)^5$:
  6. 6$1 - 4x = 0.96$
  7. 7$4x = 0.04$
  8. 8$x = 0.01$
  9. 9Substitute $x=0.01$ into the expansion:
  10. 10$(1-4(0.01))^5 \approx 1 - 20(0.01) + 160(0.01)^2 - 640(0.01)^3$
  11. 11$\approx 1 - 0.2 + 0.016 - 0.00064$
  12. 12$\approx 0.81536$
  13. 13$(0.96)^5 \approx 0.815$ (3 d.p.)
Answer:First four terms: $1 - 20x + 160x^2 - 640x^3 + \ldots$; $(0.96)^5 \approx 0.815$

Key Formulas to Remember

Binomial Theorem Formula
$(a+b)^n = a^n + \binom{n}{1}a^{n-1}b + \binom{n}{2}a^{n-2}b^2 + \ldots + \binom{n}{r}a^{n-r}b^r + \ldots + b^n$

Note: This formula is given in the Additional Mathematics Formulae Sheet during GCE O Level Examination

Binomial Coefficient
$\binom{n}{r} = \frac{n!}{r!(n-r)!} = \frac{n(n-1)\ldots(n-r+1)}{r!}$

Note: Also written as $^nC_r$

General Term
$T_{r+1} = \binom{n}{r} \times a^{(n-r)} \times b^r$

Note: For the expansion of $(a+b)^n$, the $(r+1)$th term

Special Identities to Memorize
  • $0! = 1$
  • $\binom{n}{0} = 1$
  • $\binom{n}{1} = n$
  • $\binom{n}{2} = \frac{n(n-1)}{2!}$
  • $\binom{n}{3} = \frac{n(n-1)(n-2)}{3!}$
  • $\binom{n}{n} = 1$

Ready to Master Additional Math?

Sign up for our Free Mini-Courses and try our well-structured curriculum to see how it can help maximize your learning in mathematics online.

Sign Up for Free Mini-CourseView A Math All-Access Course

Continue Your Learning

Solving Linear Equations
View Study Guide
Indices
View Study Guide
Surds
View Study Guide
Back to All Study Guides