Binomial Expansion

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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    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

    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:

    \[ \begin{align} \binom{5}{3} &= \frac{5!}{3!(5-3)!} \\ &= \frac{5\cdot 4\cdot 3 \cdot 2 \cdot 1}{(3\cdot 2\cdot 1)(2\cdot 1)} \\ &= 10. \end{align}\]

    How do you do a binomial expansion?

    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.

    Binomial expansion formula

    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.

    Binomial expansion for fractional and negative powers

    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

    \[ \sqrt{1-2x} = 1 - x - \frac{x^2}{2} - \frac{x^3}{2}.\]

    Binomial expansion questions

    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 \(1, -18x, \text{and } 162x^2\).

    Binomial Expansion - Key takeaways

    • A binomial expansion helps us to simplify algebraic expressions into a sum
    • The formula for the binomial expansion is:

      \[ (x+y)^n = \binom{n}{0}x^ny^0 + \binom{n}{1}x^{n-1}y^1 + \binom{n}{2}x^{n-2}y^2 + \dots + \binom{n}{n-1}x^1y^{n-1} + \binom{n}{n}x^0y^n\]

    • The binomial coefficients or constant terms in this expression are found using:\[ \binom{n}{k} = \frac{n!}{k!(n-k)!}\]
    • \[ (1+a)^n = 1 + na+ \frac{n(n-1)}{2!}a^2 + \frac{n(n-1)(n-2)}{3!}a^3 + \dots \]

    Frequently Asked Questions about Binomial Expansion

    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

    Test your knowledge with multiple choice flashcards

    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 \(n!\)?

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