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Find the first three terms, in ascending powers of x, of the binomial expansion of \( \frac{1}{\sqrt{4-x}} \) giving each coefficient in its simplest form - Edexcel - A-Level Maths Pure - Question 6 - 2019 - Paper 2
Question 6
Find the first three terms, in ascending powers of x, of the binomial expansion of \( \frac{1}{\sqrt{4-x}} \) giving each coefficient in its simplest form. The exp... show full transcript
Worked Solution & Example Answer:Find the first three terms, in ascending powers of x, of the binomial expansion of \( \frac{1}{\sqrt{4-x}} \) giving each coefficient in its simplest form - Edexcel - A-Level Maths Pure - Question 6 - 2019 - Paper 2
Step 1
Find the first three terms, in ascending powers of x, of the binomial expansion of \( \frac{1}{\sqrt{4-x}} \)
Answer
To find the binomial expansion, we rewrite ( \frac{1}{\sqrt{4-x}} ) as ( (4-x)^{-\frac{1}{2}} ). Using the binomial expansion formula ( (1 + u)^n = 1 + nu + \frac{n(n-1)}{2!}u^2 + ... ), where ( u = -\frac{x}{4} ) and ( n = -\frac{1}{2} ), we have:
- The first term is ( 1 ).
- The second term is ( -\frac{1}{2} \left( -\frac{x}{4} \right) = \frac{x}{8} ).
- The third term is ( \frac{-\frac{1}{2} \cdot -\frac{3}{2}}{2} \left( -\frac{x}{4} \right)^2 = \frac{3}{128} x^2 ).
So the first three terms are:
[ 1 + \frac{x}{8} + \frac{3}{128} x^2 ]
Step 2
Without evaluating your expansion, (i) state, giving a reason, which of the three values of x should not be used
Answer
The value ( x = -14 ) should not be used because substituting it makes the expression ( \sqrt{4 - (-14)} = \sqrt{18} ), which is more than 4, leading to an invalid expansion since the binomial series converges only if ( |u| < 1 ). In this case, ( |-\frac{-14}{4}| > 1 ).
Step 3
Without evaluating your expansion, (ii) state, giving a reason, which of the three values of x would lead to the most accurate approximation to \( \sqrt{2} \)
Answer
The value ( x = -\frac{1}{2} ) would lead to the most accurate approximation because it results in the term converging to the nearest point to zero, allowing for a better approximation. This leads to the expansion being valid as ( |u| < 1 ), since ( |\frac{1/2}{4}| = \frac{1}{8} < 1 ).