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Find the first three terms, in ascending powers of $x$, in the binomial expansion of \( (1 + 6x)^{\frac{1}{3}} \) - AQA - A-Level Maths Pure - Question 5 - 2019 - Paper 3
Question 5
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 approx... show full transcript
Worked Solution & Example Answer:Find the first three terms, in ascending powers of $x$, in the binomial expansion of \( (1 + 6x)^{\frac{1}{3}} \) - AQA - A-Level Maths Pure - Question 5 - 2019 - Paper 3
Step 1
Find the first three terms, in ascending powers of $x$, in the binomial expansion of \( (1 + 6x)^{\frac{1}{3}} \)
Answer
To find the first three terms of the binomial expansion of ( (1 + 6x)^{\frac{1}{3}} ), we can use the binomial theorem:
[ (a + b)^n = \sum_{k=0}^{n} \binom{n}{k} a^{n-k} b^k ]
For our expression:
- ( n = \frac{1}{3} )
- ( a = 1 )
- ( b = 6x )
Thus,
[ (1 + 6x)^{\frac{1}{3}} = \sum_{k=0}^{2} \binom{\frac{1}{3}}{k} (1)^{\frac{1}{3}-k} (6x)^k = \binom{\frac{1}{3}}{0} (1)^{\frac{1}{3}} (6x)^0 + \binom{\frac{1}{3}}{1} (1)^{\frac{-2}{3}} (6x)^1 + \binom{\frac{1}{3}}{2} (1)^{\frac{-5}{3}} (6x)^2 ]
Calculating the first three terms:
- For ( k=0 ): ( \binom{\frac{1}{3}}{0} = 1 ) gives the term ( 1 ).
- For ( k=1 ): ( \binom{\frac{1}{3}}{1} = \frac{1}{3} ) gives the term ( \frac{1}{3} \cdot 6x = 2x ).
- For ( k=2 ): ( \binom{\frac{1}{3}}{2} = \frac{\frac{1}{3}(\frac{1}{3}-1)}{2!} = -\frac{1}{9} ) gives the term ( -\frac{1}{9} \cdot (6x)^2 = -\frac{1}{9} \cdot 36x^2 = -4x^2 ).
Thus, the first three terms are:
[ 1 + 2x - 4x^2 ]
Step 2
Use the result from part (a) to obtain an approximation to \( \sqrt{1.18} \) giving your answer to 4 decimal places.
Answer
From part (a), we use the simplified expression ( (1 + 6x)^{\frac{1}{3}} \approx 1 + 2x - 4x^2 ).
To approximate ( \sqrt{1.18} ):
- We set ( 1 + 6x = 1.18 ) so ( 6x = 0.18 ) which gives ( x = 0.03 ).
Substituting ( x = 0.03 ) into the expression:
[ 1 + 2(0.03) - 4(0.03)^2 = 1 + 0.06 - 4(0.0009) ] [ = 1 + 0.06 - 0.0036 = 1.0564 ]
Thus, an approximation to ( \sqrt{1.18} ) is ( 1.0564 ).
Step 3
Explain why substituting \( x = \frac{1}{2} \) into your answer to part (a) does not lead to a valid approximation for \( \sqrt{4} \).
Answer
Substituting ( x = \frac{1}{2} ) into the approximation from part (a) does not yield a valid result because the expansion is valid only for values of ( |6x| < 1 ). Here, ( 6x = 6 \cdot \frac{1}{2} = 3 ), which exceeds 1. This means the approximation from the binomial expansion is not applicable, as the series converges only within the specified range. Therefore, using ( x = \frac{1}{2} ) would not provide a valid approximation for ( \sqrt{4} ) (which is exactly 2).