Expand $$\frac{1}{(2-5x)^2}$$ in ascending powers of $x$, up to and including the term in $x^2$, giving each term as a simplified fraction. (5) Given that the binomial expansion of $$\frac{2+kx}{(2-5x)^2}$$ is $$\frac{1}{2} + \frac{7}{4} + Ax^2 + ...$$ where $|x| < \frac{2}{5}$, (b) find the value of the constant $k$; (c) find the value of the constant $A$.

A-Level Edexcel Maths Pure Parametric Equations: Expand $$\frac{1}{(2-5x)^2}$$

Expand $$\frac{1}{(2-5x)^2}$$ in ascending powers of $x$, up to and including the term in $x^2$, giving each term as a simplified fraction - Edexcel - A-Level Maths Pure - Question 5 - 2012 - Paper 8

Question 5

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Expand $$\frac{1}{(2-5x)^2}$$ in ascending powers of $x$, up to and including the term in $x^2$, giving each term as a simplified fraction. (5) Given that the bi... show full transcript

Worked Solution & Example Answer:Expand $$\frac{1}{(2-5x)^2}$$ in ascending powers of $x$, up to and including the term in $x^2$, giving each term as a simplified fraction - Edexcel - A-Level Maths Pure - Question 5 - 2012 - Paper 8

Step 1

Expand $$\frac{1}{(2-5x)^2}$$

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Answer

To expand $\frac{1}{(2-5x)^2}$ , we can use the binomial series expansion. The binomial theorem states:

$(1 + u)^n = \sum_{k=0}^{\infty} \binom{n}{k} u^k$ for $|u| < 1$ .

For our case, we rewrite it as:

$\frac{1}{(2-5x)^2} = \frac{1}{4} \cdot \frac{1}{(1 - \frac{5x}{2})^2}$ .

This gives us:

$\frac{1}{4} \sum_{k=0}^{\infty} \binom{k+1}{1} \left(\frac{5x}{2}\right)^k \\ = \frac{1}{4} \left( 1 + 5x + \frac{(5x)^2}{2^2} + ... \right)$

Which simplifies to:

$= \frac{1}{4} \left( 1 + 5x + \frac{25x^2}{4} + ... \right)$

So up to the term in $x^2$ , we have:

$= \frac{1}{4} + \frac{5}{4}x + \frac{25}{16}x^2$ .

Step 2

Find the value of the constant $k$

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Answer

From the expansion given: $\frac{2 + kx}{(2-5x)^2} = \frac{1}{2} + \frac{7}{4} + Ax^2 + ...$

Using the earlier expansion, we have:

$2 + kx = \left( \frac{1}{2} + \frac{7}{4} \right)(2 - 5x)^2$

Evaluating for $x$ gives:

$k = -3$ .

Step 3

Find the value of the constant $A$

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Answer

To find the constant $A$ , we will expand:

$(2 + kx) \cdot \left( \frac{1}{4} + \frac{5}{4}x + \frac{25}{16}x^2 \right)$

Substituting $k = -3$ :

$= (2 - 3x) \cdot \left( \frac{1}{4} + \frac{5}{4}x + \frac{25}{16}x^2 \right)$

Multiplying out the terms gives:

$= \frac{1}{2} - \frac{3}{4}x + \frac{7}{4}x^2 + ...$

Equating coefficients of $x^2$ :

$0 = \frac{25}{8}$ implies $A$ is either $\frac{45}{8}$ or $5.625$ ; hence:

$A = 0$ .

Expand $$\frac{1}{(2-5x)^2}$$ in ascending powers of $x$, up to and including the term in $x^2$, giving each term as a simplified fraction - Edexcel - A-Level Maths Pure - Question 5 - 2012 - Paper 8

Worked Solution & Example Answer:Expand $$\frac{1}{(2-5x)^2}$$ in ascending powers of $x$, up to and including the term in $x^2$, giving each term as a simplified fraction - Edexcel - A-Level Maths Pure - Question 5 - 2012 - Paper 8

Expand $$\frac{1}{(2-5x)^2}$$

Find the value of the constant $k$

Find the value of the constant $A$

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