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4. (a) 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

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4.-(a)-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.png

4. (a) 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 fo... show full transcript

Worked Solution & Example Answer:4. (a) 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 \[ \frac{1}{\sqrt{4 - x}} \]

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Answer

To find the first three terms of the binomial expansion, we can rewrite ( \frac{1}{\sqrt{4 - x}} ) as ( (4 - x)^{-\frac{1}{2}} ). Using the binomial expansion formula ( (1 + u)^n ), we set ( u = -\frac{x}{4} ) and ( n = -\frac{1}{2} ).

Now, applying the binomial theorem:

[ (1 + u)^n = 1 + nu + \frac{n(n-1)}{2!}u^2 + \ldots ]

We get:

  1. First term: ( 1 )
  2. Second term: (-\frac{1}{2} \left( -\frac{x}{4} \right) = \frac{x}{8} )
  3. Third term: ( \frac{-\frac{1}{2} \left(-\frac{3}{2}\right)}{2} \left(-\frac{x}{4}\right)^2 = \frac{3x^2}{128} )

Thus, the first three terms of the expansion are: [ 1 + \frac{x}{8} + \frac{3x^2}{128} ]

Step 2

state, giving a reason, which of the three values of \( x \) should not be used

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Answer

x=14x = -14 should not be used because the expansion is only valid for (|u| < 1), leading to the condition (|-\frac{x}{4}| < 1), or equivalently, (-4 < x < 4). Since (x = -14) is outside this range, it would render the expansion invalid.

Step 3

state, giving a reason, which of the three values of \( x \) would lead to the most accurate approximation to \( \sqrt{2} \)

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Answer

x=2x = 2 would lead to the most accurate approximation because it results in the expansion converging exactly to ( \sqrt{2} ), while the other values would introduce inaccuracies.

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