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

## 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

$(1 - 2x)^{\frac{1}{2}} = 1 + \frac{1}{2}( - 2x) + \frac{{\frac{1}{2}\left( { - \frac{1}{2}} \right)}}{2}( - 2x)^2 + \frac{{\left( {\frac{1}{2}} \right)\left( { - \frac{1}{2}} \right)\left( { - \frac{3}{2}} \right)}}{{3 \times 2}}( - 2x)^3 + ...$

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

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 $\sqrt {98} = 10\left( {1 - 2x} \right)^{\frac{1}{2}}$

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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