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Binomial Expansion & Rational Powers (A2 only)

Maths: Edexcel A Level Pure Maths, topic 4.1.4

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.

Binomial expansion

Binomial expansion

  • The binomial expansion of a the expression (1+x)n(1+x)^n is equal to:
  • 1+nx+n(n−1)2!x2+n(n−1)(n−2)3!x3+...+n(n−1)×...×(n−r+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 n∈n\in ℝ (is a real number).
  • This equation is included in the formula booklet in the exam.
Binomial expansion

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(n−1)2!(bx)2+n(n−1)(n−2)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∣<1∣b∣|x|<\frac{\small 1}{\small |b|}.
Example

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?
Comparing variables

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.
First term

First term

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

Second term

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

Third term

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

Fourth term

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

Answer

  • The first four terms of the binomial expansion of (1+x)−2(1+x)^{-2} are:
  • (1+x)−2=1−2x+3x2−4x3+...(1+x)^{-2} = 1 -2x + 3x^2 - 4x^3 + ...
Validity

Validity

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

Test yourself on Binomial Expansion & Rational Powers (A2 only)

  • For which values of xx is the binomial expansion of 11+2x\frac{1}{1+2x} valid?

  • For which values of xx is the binomial expansion of (1+13x)−4(1+\frac{1}{3}x)^{-4} valid?

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