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CIE A-Level Maths Study Notes

2.1.5 Expansion of (1 + x)^n

The expansion of (1+x)n(1 + x)^n, where nn is a rational number and x1x \geq -1, is a crucial concept in algebra. This section covers the method of expanding such expressions, finding the general term in an expansion, adapting the standard series for various expressions, and determining the valid range of xx values for these expansions.

Binomial Series Expansion

When expanding (1+x)n(1+x)^nwhere |x| < 1, the series can be expressed as:

1+n1x+n(n1)1×2x2+n(n1)(n2)1×2×3x3+1 + \frac{n}{1}x + \frac{n(n-1)}{1 \times 2}x^2 + \frac{n(n-1)(n-2)}{1 \times 2 \times 3}x^3 + \ldots

Key Techniques for Expansion

1. Factor Case: If the constant in the brackets is not 1, extract a factor from the brackets to normalise it to 1 and then apply the general equation. Remember to adjust the indices appropriately.

2. Substitution Case: When the bracket contains more than one xx term (e.g., (2x+x2))(2 - x + x^2) ), let the complex part be u u, expand using the binomial series, and then substitute back in.

3. Determining the Limit of xx in Expansion: For an expression like (1+ax)n(1 + ax)^n, the limit of xx can be ascertained by substituting axax in the condition |x| < 1 and reformulating it to isolate xx within the modulus sign.

Practical Example

Problem: Show that, for small values of x2x^2,

(12x2)2(1+6x2)23kx4\left(1 - 2 x^2\right)^{-2} - \left(1 + 6 x^2\right)^{\frac{2}{3}} \approx k x^4

Solution:

Expanding (12x2)2(1 - 2 x^2)^{-2} up to the x4x^4 term:

1+x)2=12x+2((2)1)1×2x21 + x)^{-2} = 1 - 2x + \frac{-2((-2) - 1)}{1 \times 2}x^2=12x+3x2= 1 - 2x + 3x^2(12x2)2=14x2+12x4(1 - 2 x^2)^{-2} = 1 - 4x^2 + 12x^4

Expanding (1+6x2)23(1 + 6 x^2)^{\frac{2}{3}} up to the x4x^4 term:

(1+x)23=1+(23)x+23((23)1)1×2x2(1 + x)^{\frac{2}{3}} = 1 + \left(\frac{2}{3}\right)x + \frac{\frac{2}{3}\left(\left(\frac{2}{3}\right) - 1\right)}{1 \times 2}x^2 =1+23x19x2= 1 + \frac{2}{3}x - \frac{1}{9}x^2(1+6x2)23=1+4x24x4(1 + 6 x^2)^{\frac{2}{3}} = 1 + 4x^2 - 4x^4

Subtracting the expansions:

(14x2+12x4)(1+4x24x4)=8x2+16x4\left(1 - 4 x^2 + 12 x^4\right) - \left(1 + 4 x^2 - 4 x^4\right) = -8 x^2 + 16 x^4

The value of kk is: 16

Dr Rahil Sachak-Patwa avatar
Written by: Dr Rahil Sachak-Patwa
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Oxford University - PhD Mathematics

Rahil spent ten years working as private tutor, teaching students for GCSEs, A-Levels, and university admissions. During his PhD he published papers on modelling infectious disease epidemics and was a tutor to undergraduate and masters students for mathematics courses.

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