A binomial expansion is a method used to allow us to expand and simplify algebraic expressions in the form \( (x+y)^n\) into a sum of terms of the form \(ax^by^c\). If \(n\) is an integer, \(b\) and \(c\) also will be integers, and \(b + c = n\).
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Sign up for freeWhich of these is the correct expansion of \((x^2-y)^2\)?
Given that the coefficient of \(x^2\) in the expansion of \((1+ax)^7\) is \(525\), find the possible values of \(a\).
Which of these is equivalent to \(n!\)?
Which of these is the correct expansion of \(6 choose 2\)?
How do you read the notation \(n \choose k\)?
What does the notation \(n\choose k\) mean?
What is 6 choose 6?
What is 5 choose 0?
Which of these is the correct expansion of \((x^2-y)^2\)?
Given that the coefficient of \(x^2\) in the expansion of \((1+ax)^7\) is \(525\), find the possible values of \(a\).
Which of these is equivalent to \(n!\)?
Which of these is the correct expansion of \(6 choose 2\)?
How do you read the notation \(n \choose k\)?
What does the notation \(n\choose k\) mean?
What is 6 choose 6?
What is 5 choose 0?
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Team Binomial Expansion Teachers
We can expand expressions in the form \( (x+y)^n\) by multiplying out every single bracket, but this might be very long and tedious for high values of \(n\) such as in \( (x+y)^{20}\) for example. This is where using the Binomial Theorem comes in useful.
The binomial theorem allows us to expand an expression of the form \( (x+y)^n\) into a sum. A general formula for a binomial expression is:
\[ (x+y)^n = \binom{n}{0}x^ny^0 + \binom{n}{1}x^{n-1}y^1 + \binom{n}{2}x^{n-2}y^2 + \dots + \binom{n}{n-1}x^1y^{n-1} + \binom{n}{n}x^0y^n.\]
Which can be simplified to:
\[ \begin{align} (x+y)^n &= \sum\limits_{k=0}^n \binom{n}{k} x^{n-k}y^k \\ &= \sum\limits_{k=0}^n \binom{n}{k} x^ky^{n-k} . \end{align}\]
Where both \(n\) and \(k\) are integers. This is also known as the binomial formula. The notation
\[ \binom{n}{k}\]
can be referred to as '\(n\) choose \(k\)' and gives a number called the binomial coefficient which is the number of different combinations of ordering \(k\) objects out of a total of \(n\) objects. The equation for the binomial coefficient (\(n\) choose \(k\) or \(^nC_r\) on a calculator) is given by:
\[ \binom{n}{k} = \frac{n!}{k!(n-k)!}\]
Where '!' means factorial. Factorial means the product of an integer with all the integers below it. For example for \(5\) choose \(3\), we would have:
\[ \begin{align} \binom{5}{3} &= \frac{5!}{3!(5-3)!} \\ &= \frac{5\cdot 4\cdot 3 \cdot 2 \cdot 1}{(3\cdot 2\cdot 1)(2\cdot 1)} \\ &= 10. \end{align}\]
To understand how to do a binomial expansion, we will look at an example. Let's say we want to expand \( (x+y)^4\). In this case, \(n = 4\) and \(k\) will vary between \(0\) and \(4\). Using the formula for the binomial expansion, we can write:
\[ (x+y)^4 = \binom{4}{0}x^4y^0 + \binom{4}{1}x^3y^1 + \binom{4}{2}x^2y^2 + \binom{4}{3}x^1y^3+\binom{4}{4}x^0y^4.\]
We can now use the equation for the binomial coefficient to find all the constant terms in this expression. For the first term we have:
\[ \begin{align} \binom{4}{0} &= \frac{4!}{0!(4-0)!} \\ &= \frac{4 \cdot 3\cdot 2\cdot 1}{1\cdot (4 \cdot 3\cdot 2\cdot 1 )} \\ &= 1. \end{align} \]
Repeating this for all five coefficients, we end up with binomial coefficients of \(1\), \(4\), \(6\), \(4\), \(1\) in order. Therefore, our expression for the binomial expansion simplifies to:
\[ x^4 + 4x^3y + 6x^2y^2 + 4xy^3 + y^4.\]
Note that \(y\) could also be replaced by any number.
To summarise the above explanation, the expansion formula can be written as:
\[(x+y)^n = \sum _{k=0}^{n} \binom{n}{k}x^{n-k}y^k = \sum _{k=0}^{n} \binom{n}{k}x^{k}y^{n-k}\]
Where \(\binom{n}{k}\) is the binomial coefficient of each term.
Sometimes you will encounter algebraic expressions where n is not a positive integer but a negative integer or a fraction. Let's consider the expression \(\sqrt{1-2x}\) which can also be written as
\[ (1- 2x)^\dfrac{1}{2} \] where \(x < 0.5\). In this case, it becomes hard to find the formula to find the binomial coefficients,
because we can't find the factorials for a negative or rational number. However, if we look at an example for a positive integer, we can find a more general expression that we can then also apply to negative and fractional numbers. For example for
\[ \binom{6}{3}\]
we have
\[ \begin{align} \binom{6}{3}&= \frac{6!}{3!(6-3)!} \\ &= \frac{6\cdot 5\cdot 4}{3!} \\ &= \frac{6(6-1)(6-2)}{3!}. \end{align}\]
From this we observe that
\[ \binom{n}{k} = \frac{n(n-1)(n-2)(n-3)\dots (n-k+1)}{k!} \]
and therefore the more general expression for the binomial theorem is the infinite formula
\[ (a+b)^n = \frac{a^n}{0!} + \frac{na^{n-1}b}{1!} + \frac{n(n-1)a^{n-2}b^2}{2!} + \frac{n(n-1)(n-2)a^{n-3}b^3}{3!} + \dots \]
Let's look at \(\sqrt{1-2x}\). In this case \(a = -2x\), \(b = 1\) and \(n =1/2\). Substituting this we get:
\[ \begin{align} \frac{(-2x)^\frac{1}{2}}{0!} &+ \frac{\left(-\frac{1}{2}\right) (-2x)^{-\frac{1}{2}}\cdot 1 }{1!} \\ &\quad + \frac{\left(-\frac{1}{2}\right) \left(-\frac{1}{2}\right) (-2x)^{-\frac{3}{2}}\cdot 1^2 }{2!} \\ &\quad + \frac{\left(-\frac{1}{2}\right) \left(-\frac{1}{2}\right) \left(-\frac{3}{2}\right) (-2x)^{-\frac{5}{2}}\cdot 1^3 }{3!} + \dots \end{align}\]
Using Mac Laurin's expansion we can say that the above expression converges to
\[ \sqrt{1-2x} = 1 - x - \frac{x^2}{2} - \frac{x^3}{2}.\]
We have collected a few questions with step-by-step solutions to help you understand how the binomial theorem and binomial expansion can be applied or asked about in an exam.
Exercise 1
Expand \((x + 2)^4\) using the binomial theorem.
Solution:
Using the binomial theorem, we have:
\((x + 2)^4 = \binom{4}{0}x^4(2)^0 + \binom {4}{1}x^3(2)^1 + \binom{4}{2}x^2(2)^2 + \binom{4}{3}x(2)^3 + \binom{4}{4}(2)^4\)
Evaluating the coefficients, we get:
\((x + 2)^4 = x^4 + 8x^3 + 24x^2 + 32x + 16\)
Therefore, \((x + 2)^4\) expands to \(x^4 + 8x^3 + 24x^2 + 32x + 16\).
Exercise 2
Find the coefficient of \(x^3\) in the expansion of \((2x + 1)^5\).
Solution:
Using the binomial theorem, the expansion of \((2x + 1)^5\) is:
\((2x + 1)^5 = \binom{5}{0}(2x)^0(1)^5 + \binom{5}{1}(2x)^1(1)^4 + \binom{5}{2}(2x)^2(1)^3 + \binom{5}{3}(2x)^3(1)^2 + \binom{5}{4}(2x)^4(1)^1 + \binom{5}{5}(2x)^5(1)^0\)
To find the coefficient of \(x^3\), we need to look at the term with \((2x)^3\):
\(\binom{5}{3}(2x)^3(1)^2 = 10(2x)^3\)
Evaluating the term, we get:
\(10(2x)^3 = 80x^3\)
Therefore, the coefficient of \(x^3\) in the expansion of \((2x + 1)^5\) is 80.
Exercise 3
Find the first three terms in the expansion of \((1 - 3x)^6\).
Solution:
Using the binomial theorem, the expansion of \((1 - 3x)^6\) is:
\((1 - 3x)^6 = \binom{6}{0}(1)^6(-3x)^0 + \binom{6}{1}(1)^5(-3x)^1 + \binom{6}{2}(1)^4(-3x)^2 + ...\)
To find the first three terms, we need to evaluate the terms with \((1)^6, (1)^5, \text{and} \space (1)^4\):
\(\binom {6}{0}(1)^6(-3x)^0 = 1\)
\(\binom{6}{1}(1)^5(-3x)^1 = -18x\)
\(\binom{6}{2}(1)^4(-3x)^2 = 162x^2\)
Therefore, the first three terms in the expansion of \((1 - 3x)^6\) are \(1, -18x, \text{and } 162x^2\).
The formula for the binomial expansion is:
\[ (x+y)^n = \binom{n}{0}x^ny^0 + \binom{n}{1}x^{n-1}y^1 + \binom{n}{2}x^{n-2}y^2 + \dots + \binom{n}{n-1}x^1y^{n-1} + \binom{n}{n}x^0y^n\]
To solve a binomial expansion with negative or fractional exponents, we use:
\[ (1+a)^n = 1 + na+ \frac{n(n-1)}{2!}a^2 + \frac{n(n-1)(n-2)}{3!}a^3 + \dots \]
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How do you find the constant term in a binomial expansion?
The constant term is found by using the formula
n choose k=n!/k!(n-k)!
What is binomial expansion?
A binomial expansion is a method that allows us to simplify complex algebraic expressions into a sum.
How do you do a binomial expansion?
You can use the binomial expansion formula
(x+y)^n=(nC0)x^n y^0+(nC1)x^/n-1)y^1+(nC2)x^(n-2)y^2+...+(nCn-1)x^1y^(n-1)+(nCn)x^0y^n
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