1. 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 coefficient in its simplest form. (b) Use the expansion from part (a), with a suitable value of $x$, to find an approximate value for $\sqrt{ \frac{3}{10} }$. Show all your working and give your answer to 3 decimal places.

A-Level Edexcel Maths Pure Implicit Differentiation: 1. Find the binomial series expa

1. 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 2 - 2018 - Paper 9

Question 2

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1. 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 coef... show full transcript

Worked Solution & Example Answer:1. 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 2 - 2018 - Paper 9

Step 1

Find the binomial series expansion of $$\sqrt{4 - 9x}, \, |x| < \frac{4}{9}$$

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Answer

To find the binomial series expansion, we can rewrite the expression as:

$\sqrt{4 - 9x} = (4(1 - \frac{9}{4}x))^{1/2} = 2(1 - \frac{9}{4}x)^{1/2}$

Now we can apply the binomial series expansion, which is given by:

$(1 + u)^n \approx 1 + nu + \frac{n(n-1)}{2!}u^2 + \ldots$

In this case, $n = \frac{1}{2}$ and $u = -\frac{9}{4}x$ :

First term: $1$
Second term: $\frac{1}{2} \left(-\frac{9}{4}x\right) = -\frac{9}{8}x$
Third term: $\frac{\frac{1}{2}(\frac{1}{2} - 1)}{2!}\left(-\frac{9}{4}x\right)^2 = \frac{\frac{1}{2}(-\frac{1}{2})}{2}\left(\frac{81}{16}x^2\right) = -\frac{81}{64}x^2$

Putting it all together:

$\sqrt{4 - 9x} = 2 \left(1 - \frac{9}{8}x - \frac{81}{64}x^2 + \ldots\right) = 2 - \frac{9}{4}x - \frac{81}{32}x^2 + \ldots$

Thus, the expansion up to the term in $x^2$ is: $2 - \frac{9}{4}x - \frac{81}{32}x^2$ .

Step 2

Use the expansion from part (a), with a suitable value of $x$, to find an approximate value for $\sqrt{ \frac{3}{10} }$.

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Answer

We want to approximate $\sqrt{\frac{3}{10}}$ . We can express $\frac{3}{10}$ in a suitable form:

$\frac{3}{10} = \frac{3}{4} \cdot \frac{4}{10} = \frac{3}{4} \cdot \frac{2}{5}$

Now to use the expansion from part (a), we let:

$4 - 9x = \frac{3}{10}$

This requires setting $x$ such that:

$4 - 9x = \frac{3}{10} \implies 9x = 4 - \frac{3}{10} = \frac{40}{10} - \frac{3}{10} = \frac{37}{10} \implies x = \frac{37}{90}$

Now substituting this value into our expansion, we have:

$\sqrt{\frac{3}{10}} \approx 2 - \frac{9}{4}\left(\frac{37}{90}\right) - \frac{81}{32}\left(\frac{37}{90}\right)^2$

Calculating each term:

First term: $2 = 2$
Second term: $-\frac{9 \cdot 37}{4 \cdot 90} = -\frac{333}{360} \approx -0.925$
Third term:

$-\frac{81 \cdot 1369}{32 \cdot 8100} \approx -0.195$

Adding these results together gives:

$\sqrt{\frac{3}{10}} \approx 2 - 0.925 - 0.195 = 0.880$

Rounding to three decimal places, we find:

$\sqrt{\frac{3}{10}} \approx 0.880$ .

1. 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 2 - 2018 - Paper 9

Worked Solution & Example Answer:1. 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 2 - 2018 - Paper 9

Find the binomial series expansion of $$\sqrt{4 - 9x}, \, |x| < \frac{4}{9}$$

Use the expansion from part (a), with a suitable value of $x$, to find an approximate value for $\sqrt{ \frac{3}{10} }$.

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