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Find the binomial series expansion of $$ \sqrt{4 - 9x}, \ |x| < \frac{4}{9} $$ in ascending powers of $x$, up to and including the term in $x^2$ - Edexcel - A-Level Maths Pure - Question 3 - 2018 - Paper 9

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Find-the-binomial-series-expansion-of-$$-\sqrt{4---9x},-\-|x|-<-\frac{4}{9}--$$-in-ascending-powers-of-$x$,-up-to-and-including-the-term-in-$x^2$-Edexcel-A-Level Maths Pure-Question 3-2018-Paper 9.png

Find the binomial series expansion of $$ \sqrt{4 - 9x}, \ |x| < \frac{4}{9} $$ in ascending powers of $x$, up to and including the term in $x^2$. Give each coeffici... show full transcript

Worked Solution & Example Answer:Find the binomial series expansion of $$ \sqrt{4 - 9x}, \ |x| < \frac{4}{9} $$ in ascending powers of $x$, up to and including the term in $x^2$ - Edexcel - A-Level Maths Pure - Question 3 - 2018 - Paper 9

Step 1

Find the binomial series expansion of $\sqrt{4 - 9x}$

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Answer

To find the binomial series expansion, we start by rewriting the expression:

49x=(4(194x))1/2=2194x.\sqrt{4 - 9x} = (4(1 - \frac{9}{4}x))^{1/2} = 2\sqrt{1 - \frac{9}{4}x}.

Using the binomial expansion for (1+u)n(1 + u)^n, where n=12n = \frac{1}{2} and u=94xu = -\frac{9}{4}x, we expand:

(1u)1/2=112u+12(121)u22!+(1 - u)^{1/2} = 1 - \frac{1}{2}u + \frac{1}{2}\left(\frac{1}{2}-1\right)\frac{u^2}{2!} + \ldots

Substituting our expression gives us:

112(94x)+12(121)(94x)221 - \frac{1}{2}\left(-\frac{9}{4}x\right) + \frac{1}{2}\left(\frac{1}{2}-1\right)\frac{(-\frac{9}{4}x)^2}{2}

Calculating each term:

  • First term: 11
  • Second term: 98x\frac{9}{8}x
  • Third term: 8164x22=81128x2-\frac{81}{64} \cdot \frac{x^2}{2} = -\frac{81}{128}x^2

Putting it together:

49x2(1+98x81128x2)=2+94x8164x2.\sqrt{4 - 9x} \approx 2\left(1 + \frac{9}{8}x - \frac{81}{128}x^2\right) = 2 + \frac{9}{4}x - \frac{81}{64}x^2.

Thus, the final expansion up to x2x^2 is:

49x2+94x8164x2.\sqrt{4 - 9x} \approx 2 + \frac{9}{4}x - \frac{81}{64}x^2.

Step 2

Use the expansion from part (a), with a suitable value of $x$, to find an approximate value for $\sqrt{310}$

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Answer

To approximate 310\sqrt{310}, we first note that 310310 can be expressed in terms of 49x4 - 9x:

Set 49x=3104 - 9x = 310, which gives us:

However, finding a suitable xx in the acceptable range x<490.444|x| < \frac{4}{9} \approx 0.444 is essential. We can rewrite 310310 as:

31049(18 (for example)).\sqrt{310} \approx \sqrt{4 - 9\left(-\frac{1}{8} \text{ (for example)} \right)}.

Using this value,

  • Calculate for x=18x = \frac{-1}{8}:

From the expansion:

49(18)2+94(18)8164(18)2.\sqrt{4 - 9(-\frac{1}{8})} \approx 2 + \frac{9}{4}(-\frac{1}{8}) - \frac{81}{64}(-\frac{1}{8})^2.

Evaluating terms:

  • Second term: 932\frac{-9}{32}
  • Third term: 816464=81512\frac{81}{64 \cdot 64} = \frac{81}{512}

Combine:

ApproximateValue2932+8151217.623 (to 3 decimal places).Approximate Value \approx 2 - \frac{9}{32} + \frac{81}{512} \approx 17.623\ (to\ 3\ decimal\ places).

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