4.1.3

Binomial Expansion

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Binomial Expansion

We can work out the expansion of binomials using an equation involving factorials.

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Expanding binomials

  • We have seen how we can expand binomials using the numbers in each row of Pascal's triangle.
  • We've also seen how to work out the numbers in Pascal's triangle using factorials.
  • This leads us to an equation we can use to expand binomials without having to think about Pascal's triangle.
    • This is called the binomial expansion or binomial theorem.
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Binomial expansion equation

  • The binomial expansion of (a+b)n(a+b)^n is:
    • (a+b)n=an+(n1)an1b+(n2)an2b2+...+(nr)anrbr+...+bn(a+b)^n=a^n + {n\choose 1}a^{n-1}b + {n\choose 2}a^{n-2}b^2 + ...+{n \choose r}a^{n-r}b^{r}+...+ b^n
  • This equation applies when nn \in (is a positive whole number).
  • This equation is included in the formula booklet in the exam.
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General term

  • The general term in a binomial expansion is given by (nr)anrbr{n \choose r}a^{n-r}b^{r}.
  • We can use this to find coefficients of specific orders of variables in the binomial expansion.
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Example

  • Use the binomial theorem to find the expansion of (16x)5(1-6x)^5.
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Comparing variables

  • If we compare the expression we want to expand with the equation given in the formula booklet:
    • a=1,b=6xa = 1, b = -6x and n=5n = 5.
  • So we can substitute these into each term to find the binomial expansion.
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First term

  • The first term of the binomial expansion is ana^n.
  • Substituting in a=1a = 1 and n=5n = 5, we have:
    • an=15=1a^n = 1^5 = 1
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Second term

  • The second term of the binomial expansion is (n1)an1b{n\choose 1}a^{n-1}b.
  • Substituting in for a,ba,b and nn, we have:
    • (n1)an1b=(51)14(6x){n\choose 1}a^{n-1}b = {5\choose 1}1^4(-6x)
  • Simplifying this term gives:
    • 5!1!×4!×(6x)=30x\frac{5!}{1!\times4!}\times(-6x) = -30x
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Third term

  • The third term of the binomial expansion is (n2)an2b2{n\choose 2}a^{n-2}b^2.
  • Substituting in for a,ba,b and nn, we have:
    • (n2)an2b2=(52)13(6x)2{n\choose 2}a^{n-2}b^2={5\choose 2}1^{3}(-6x)^2
  • Simplifying this term gives:
    • 5!2!×3!×36x2=10×36x2=360x2\frac{5!}{2!\times3!}\times36x^2 = 10\times36x^2 = 360x^2
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Fourth term

  • The fourth term of the binomial expansion is (n3)an3b3{n\choose 3}a^{n-3}b^3.
  • Substituting in for a,ba,b and nn, we have:
    • (n3)an3b3=(53)12(6x)3{n\choose 3}a^{n-3}b^3={5\choose 3}1^{2}(-6x)^3
  • Simplifying this term gives:
    • 5!3!×2!×(216)x3=10×(216)x3=2160x3\frac{5!}{3!\times2!}\times(-216)x^3 = 10\times(-216)x^3 = -2160x^3
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Fifth term

  • The fifth term of the binomial expansion is (n4)an4b4{n\choose 4}a^{n-4}b^4.
  • Substituting in for a,ba,b and nn, we have:
    • (n4)an4b4=(54)ab4{n\choose 4}a^{n-4}b^4 = {5\choose 4}ab^4
  • Simplifying this term gives:
    • 5!4!×1!×1296x4=5×1296x4=6480x4\frac{5!}{4!\times1!}\times1296x^4 = 5\times1296x^4 = 6480x^4
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Final term

  • The sixth term of the binomial expansion is (n5)an5b5{n\choose 5}a^{n-5}b^5.
    • We can tell that this is the final term in the expansion because there will be no powers of aa left in the term and that (55)=1{5\choose 5}=1.
  • So this term is simply equal to bnb^n.
  • Substituting in for bb and nn, we have:
    • bn=(6x)5=7776x5b^n = (-6x)^5 = -7776x^5
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Sum the terms

  • The expansion of (16x)5(1-6x)^5 is equal to the sum of the terms in the binomial expansion, so the answer is:
    • (16x)5=130x+360x22160x3+6480x47776x5(1-6x)^5 = 1 -30x + 360x^2 -2160x^3 + 6480x^4-7776x^5

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1Proof

2Algebra & Functions

2.1Powers & Roots

2.2Quadratic Equations

2.3Inequalities

2.4Polynomials

2.5Graphs

2.6Functions

2.7Transformation of Graphs

2.8Partial Fractions (A2 Only)

3Coordinate Geometry

4Sequences & Series

5Trigonometry

6Exponentials & Logarithms

7Differentiation

8Integration

9Numerical Methods

10Vectors

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