1. (a) Find the binomial expansion of $$\frac{1}{\sqrt{9 - 10x}}$$, where $|x| < \frac{9}{10}$, in ascending powers of $x$ up to and including the term in $x^2$. Give each coefficient as a simplified fraction. (b) Hence, or otherwise, find the expansion of $$\frac{3 + x}{\sqrt{9 - 10x}}$$, where $|x| < \frac{9}{10}$, in ascending powers of $x$, up to and including the term in $x^2$. Give each coefficient as a simplified fraction.

A-Level Edexcel Maths Pure Differentiation: 1. (a) Find the binomial expansi

1. (a) Find the binomial expansion of $$\frac{1}{\sqrt{9 - 10x}}$$, where $|x| < \frac{9}{10}$, in ascending powers of $x$ up to and including the term in $x^2$ - Edexcel - A-Level Maths Pure - Question 3 - 2014 - Paper 8

Question 3

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1. (a) Find the binomial expansion of $$\frac{1}{\sqrt{9 - 10x}}$$, where $|x| < \frac{9}{10}$, in ascending powers of $x$ up to and including the term in $x^2$. G... show full transcript

Worked Solution & Example Answer:1. (a) Find the binomial expansion of $$\frac{1}{\sqrt{9 - 10x}}$$, where $|x| < \frac{9}{10}$, in ascending powers of $x$ up to and including the term in $x^2$ - Edexcel - A-Level Maths Pure - Question 3 - 2014 - Paper 8

Step 1

Find the binomial expansion of $$\frac{1}{\sqrt{9 - 10x}}$$

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Answer

To find the binomial expansion, rewrite the expression:

$\frac{1}{\sqrt{9 - 10x}} = (9 - 10x)^{-1/2}$ .

Using the binomial series expansion:

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

Here, set $u = -\frac{10x}{9}$ , so

$\frac{1}{\sqrt{9 - 10x}} = 9^{-1/2}(1 - \frac{10x}{9})^{-1/2}$ .

Using $\sqrt{9} = 3$ , we can write:

$= \frac{1}{3}(1 - \frac{10x}{9})^{-1/2}$

The expansion for $(1 - u)^{-1/2}$ starts as:

$1 + \frac{1}{2}u + \frac{3}{8}u^2 + ...$

Substituting $u$ we get:

$= 1 + \frac{1}{2}\left(-\frac{10x}{9}\right) + \frac{3}{8}\left(-\frac{10x}{9}\right)^2$

Calculating the terms:

The constant term: 1
The first term: $\frac{-5x}{9}$
The second term: $\frac{3 \cdot 100x^2}{72} = \frac{25x^2}{6}$

So, the expansion is:

$\frac{1}{3}(1 - \frac{5x}{9} + \frac{25}{6}x^2) = \frac{1}{3} - \frac{5x}{27} + \frac{25x^2}{18}$ .

Step 2

Hence, or otherwise, find the expansion of $$\frac{3 + x}{\sqrt{9 - 10x}}$$

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Answer

From part (a), we have:

$\frac{1}{\sqrt{9 - 10x}} = \frac{1}{3} - \frac{5x}{27} + \frac{25x^2}{18}$ .

To find the expansion of $\frac{3 + x}{\sqrt{9 - 10x}}$ , we multiply:

For constant term, we have:

$\frac{1}{3} \cdot 3 = 1$

The first term:

$(-\frac{5x}{27}) \cdot 3 + \frac{1}{3} \cdot x = -\frac{15x}{27} + \frac{x}{3} = -\frac{15x}{27} + \frac{9x}{27} = -\frac{6x}{27} = -\frac{2x}{9}$

For the second term, we combine:

$\frac{25x^2}{18} \cdot 3 + (-\frac{5x}{27}) \cdot x + \frac{1}{3} \cdot x^2$

Calculating gives:

$\frac{75x^2}{18} - \frac{5x^2}{27} + \frac{x^2}{3}$

Finding a common denominator (54):

$\frac{225x^2}{54} - \frac{10x^2}{54} + \frac{18x^2}{54} = \frac{233x^2}{54}$

Combining all terms, we have:

$1 - \frac{2x}{9} + \frac{233x^2}{54}$ .

1. (a) Find the binomial expansion of $$\frac{1}{\sqrt{9 - 10x}}$$, where $|x| < \frac{9}{10}$, in ascending powers of $x$ up to and including the term in $x^2$ - Edexcel - A-Level Maths Pure - Question 3 - 2014 - Paper 8

Worked Solution & Example Answer:1. (a) Find the binomial expansion of $$\frac{1}{\sqrt{9 - 10x}}$$, where $|x| < \frac{9}{10}$, in ascending powers of $x$ up to and including the term in $x^2$ - Edexcel - A-Level Maths Pure - Question 3 - 2014 - Paper 8

Find the binomial expansion of $$\frac{1}{\sqrt{9 - 10x}}$$

Hence, or otherwise, find the expansion of $$\frac{3 + x}{\sqrt{9 - 10x}}$$

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