A-Level AQA Maths Mechanics Quantities, Units & Modelling: Find the first three terms, in a

Find the first three terms, in ascending powers of $x$, in the binomial expansion of $ (1 + 6x)^{\frac{1}{3}} $ Use the result from part (a) to obtain an approximation to $ \sqrt{1.18} $ giving your answer to 4 decimal places - AQA - A-Level Maths Mechanics - Question 5 - 2019 - Paper 3

Question 5

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Find the first three terms, in ascending powers of $x$, in the binomial expansion of $ (1 + 6x)^{\frac{1}{3}} $ Use the result from part (a) to obtain an approxi... show full transcript

Worked Solution & Example Answer:Find the first three terms, in ascending powers of $x$, in the binomial expansion of $ (1 + 6x)^{\frac{1}{3}} $ Use the result from part (a) to obtain an approximation to $ \sqrt{1.18} $ giving your answer to 4 decimal places - AQA - A-Level Maths Mechanics - Question 5 - 2019 - Paper 3

Step 1

Find the first three terms, in ascending powers of $x$, in the binomial expansion of (1 + 6x)^{1/3}

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Answer

To find the first three terms in the expansion of ( (1 + 6x)^{\frac{1}{3}} ), we can use the binomial theorem, which states:

$(a + b)^n = \sum_{k=0}^{n} \binom{n}{k} a^{n-k} b^k$

In our case, ( a = 1 ), ( b = 6x ), and ( n = \frac{1}{3} ). We can calculate the first three terms (for k = 0, 1, 2):

For ( k = 0 ): ( \binom{\frac{1}{3}}{0} (1)^1 (6x)^0 = 1 )
For ( k = 1 ): ( \binom{\frac{1}{3}}{1} (1)^{\frac{1}{3}-1} (6x)^1 = \frac{1}{3}(6x) = 2x )
For ( k = 2 ): ( \binom{\frac{1}{3}}{2} (1)^{\frac{1}{3}-2} (6x)^2 = \frac{\frac{1}{3}(\frac{1}{3}-1)}{2}(36x^2) = -\frac{1}{6}(36x^2) = -6x^2 )

Putting it all together, the first three terms in the expansion are: ( 1 + 2x - 6x^2 ).

Step 2

Use the result from part (a) to obtain an approximation to $ \sqrt{1.18} $ giving your answer to 4 decimal places.

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Answer

From part (a), we have approximated ( (1 + 6x)^{\frac{1}{3}} ) for small values of ( x ). To approximate ( \sqrt{1.18} ), we can set:

Let ( 1 + 6x = 1.18 \Rightarrow 6x = 0.18 \Rightarrow x = 0.03 )
Substitute ( x = 0.03 ) into our expansion: ( (1 + 6(0.03))^{\frac{1}{3}} \approx 1 + 2(0.03) - 6(0.03)^2 )
Calculating this: ( 1 + 0.06 - 6(0.0009) \approx 1 + 0.06 - 0.0054 = 1.0546 )

Thus, ( \sqrt{1.18} \approx 1.0546 ).

Step 3

Explain why substituting $ x = -\frac{1}{2} $ into your answer to part (a) does not lead to a valid approximation for $ \sqrt{4} $.

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Answer

Substituting ( x = -\frac{1}{2} ) into our expression from part (a) would lead us to evaluate ( (1 + 6(-\frac{1}{2}))^{\frac{1}{3}} ). This simplifies to ( (1 - 3)^{\frac{1}{3}} = (-2)^{\frac{1}{3}} ).

However, the binomial expansion used in part (a) is only valid for ( |6x| < 1 ), and in this case, substituting ( x = -\frac{1}{2} ) gives ( |6(-\frac{1}{2})| = 3 ), which does not satisfy this condition. Therefore, it does not provide a valid approximation for ( \sqrt{4} ).

Find the first three terms, in ascending powers of $x$, in the binomial expansion of \( (1 + 6x)^{\frac{1}{3}} \) Use the result from part (a) to obtain an approximation to \( \sqrt{1.18} \) giving your answer to 4 decimal places - AQA - A-Level Maths Mechanics - Question 5 - 2019 - Paper 3

Find the first three terms, in ascending powers of $x$, in the binomial expansion of (1 + 6x)^{1/3}

Use the result from part (a) to obtain an approximation to \( \sqrt{1.18} \) giving your answer to 4 decimal places.

Explain why substituting \( x = -\frac{1}{2} \) into your answer to part (a) does not lead to a valid approximation for \( \sqrt{4} \).

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