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Sequences and Series : Key Stage 5 (A2-Level) : Mathematics Content Knowledge


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Key Stage 5 (A2-Level)
Sequences and Series
Question 1 of 1

1. How confident are you that you know and can use:

a. the binomial series for any rational n?


Example

(1 + x)n, where is rational can be written as a series:


1 + nx + \frac{{n(n - 1)}}{{2!}}x^2  + \frac{{n(n - 1)(n - 2)}}{{3!}}x^3  + ...

Unless n is a positive integer, this is an infinite series.
This series will converge if |n| < 1.

Example : Expand \sqrt {(1 - 2x)} up to the term in x3.

Using the expansion of (1 + x)n and substituting -2x for x and n =

1
2
 gives

(1 - 2x)^{\frac{1}{2}}  = 1 + \frac{1}{2}( - 2x) + \frac{{\frac{1}{2}\left( { - \frac{1}{2}} 
ight)}}{2}( - 2x)^2  + \frac{{\left( {\frac{1}{2}} 
ight)\left( { - \frac{1}{2}} 
ight)\left( { - \frac{3}{2}} 
ight)}}{{3 	imes 2}}( - 2x)^3  + ...

 = 1 - x - \frac{{x^2 }}{2} - \frac{{x^3 }}{2} + ...

Note: this series is only valid for |-2x| < 1 ie valid for |x| <

1
2

The binomial expansion can be used to find approximations to irrational roots.

Example:
Find to 2dp an approximation to \sqrt{98}.

\sqrt {98}  = \sqrt {100 - 2}  = \sqrt {\frac{{100}}{{100}}(100 - 2)}  = 10\sqrt {1 - \frac{2}{{100}}}

If we let x = 0.01 then \sqrt {98}  = 10\left( {1 - 2x} 
ight)^{\frac{1}{2}}

From the series expansion above (1 - 2x)^{\frac{1}{2}}  = 1 - x - \frac{{x^2 }}{2} - \frac{{x^3 }}{2} + ...

Therefore \sqrt {98}  = 10(1 - .01 - \frac{{.01^2 }}{2} - \frac{{.01^3 }}{2} + ...)

= 10 (0.989994995+..)

= 9.89994

= 9.90 to 2dp.

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