4.1.4

Binomial Expansion & Rational Powers (A2 only)

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Binomial Expansion with Rational Powers

If we have an expression (1+x)n(1+x)^n and nn is a negative or rational number, we need to use a different equation for its expansion.

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Binomial expansion

  • The binomial expansion of a the expression (1+x)n(1+x)^n is equal to:
    • 1+nx+n(n1)2!x2+n(n1)(n2)3!x3+...+n(n1)×...×(nr+1)r!xr+...1+ nx + \frac{n(n-1)}{2!}x^2 + \frac{n(n-1)(n-2)}{3!}x^3 + ... + \frac{n(n-1)\times...\times(n-r+1)}{r!}x^r+...
  • This is an infinite sum of terms, also known as an infinite series.
  • This expansion is valid when x<1|x|<1 and nn\in (is a real number).
  • This equation is included in the formula booklet in the exam.
Illustrative background for Binomial expansionIllustrative background for Binomial expansion ?? "content

Binomial expansion

  • We can use this equation to work out the expansion of binomials of the form (1+bx)n(1+bx)^n by grouping together (bx)(bx) and treating it as if it was xx:
    • (1+bx)n=1+n(bx)+n(n1)2!(bx)2+n(n1)(n2)3!(bx)3...(1+bx)^n=1 + n(bx) + \frac{n(n-1)}{2!}(bx)^2 + \frac{n(n-1)(n-2)}{3!}(bx)^3...
  • This expansion is valid when bx<1|bx|<1 or x<1b|x|<\frac{\small 1}{\small |b|}.
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Example

  • What are the first four terms of the binomial expansion of (1+x)2(1+x)^{-2}?
    • For which values of xx is the expansion valid?
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Comparing variables

  • If we compare the expression we want to expand with the equation given in the formula booklet, we see that n=2n = -2.
  • So we can substitute this into each term to find the binomial expansion.
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First term

  • The first term of this type of binomial expansion is always equal to 1.
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Second term

  • The second term of the binomial expansion is nxnx.
  • Substituting in n=2n = -2, we get:
    • nx=2xnx = -2x
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Third term

  • The third term of the binomial expansion is n(n1)2!x2\frac{n(n-1)}{2!}x^2.
  • Substituting in n=2n = -2, we get:
    • n(n1)2!x2=(2)×(3)2×1x2=3x2\frac{n(n-1)}{2!}x^2 = \frac{(-2)\times(-3)}{2\times1}x^2 = 3x^2
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Fourth term

  • The third term of the binomial expansion is n(n1)(n2)3!x3\frac{n(n-1)(n-2)}{3!}x^3.
  • Substituting in n=2n = -2, we get:
    • n(n1)(n2)3!x3=(2)×(3)×(4)3×2×1x3=4x3\frac{n(n-1)(n-2)}{3!}x^3=\frac{(-2)\times(-3)\times(-4)}{3\times2\times1}x^3 = -4x^3
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Answer

  • The first four terms of the binomial expansion of (1+x)2(1+x)^{-2} are:
  • (1+x)2=12x+3x24x3+...(1+x)^{-2} = 1 -2x + 3x^2 - 4x^3 + ...
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Validity

  • The expansion is valid for values of xx such that x|x|<1.

Jump to other topics

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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