Expand and simplify $\left(\frac{1}{2}-2x\right)^5$ in ascending powers of $x$, up to the first 4 terms. Hence find the value of $a$ if the coefficient of $x^2$ in the expansion of
$\left(1+ax+3x^2\right)\left(\frac{1}{2}-2x\right)^5 \text{ is } \frac{13}{2}.$ Using the answer from part (i), evaluate $(0.47)^5$ correct to 5 decimal places.

1. Apply the Binomial Theorem to expand the expression up to the first 4 terms

$\left(\frac{1}{2}-2x\right)^5 = \left(\frac{1}{2}\right)^5 + \binom{5}{1}\left(\frac{1}{2}\right)^4(-2x) + \binom{5}{2}\left(\frac{1}{2}\right)^3(-2x)^2 + \binom{5}{3}\left(\frac{1}{2}\right)^2(-2x)^3 + \dots$

2. Simplify each term to obtain the final expansion

$= \frac{1}{32} + 5\left(\frac{1}{16}\right)(-2x) + 10\left(\frac{1}{8}\right)(4x^2) + 10\left(\frac{1}{4}\right)(-8x^3)$
$= \frac{1}{32} - \frac{10}{16}x + \frac{40}{8}x^2 - \frac{80}{4}x^3$
$= \frac{1}{32} - \frac{5}{8}x + 5x^2 - 20x^3$

1. Expand $\left(\frac{1}{2}-2x\right)^5$ up to the $x^2$ term (from part (i)).

$\left(\frac{1}{2}-2x\right)^5 = \left(\frac{1}{2}\right)^5 + \binom{5}{1}\left(\frac{1}{2}\right)^4(-2x) + \binom{5}{2}\left(\frac{1}{2}\right)^3(-2x)^2 + \dots$
$= \frac{1}{32} - \frac{5}{8}x + 5x^2 + \dots$

2. Identify the terms in the product that result in $x^2$.

In the product $\left(1+ax+3x^2\right)\left(\frac{1}{32} - \frac{5}{8}x + 5x^2 + \dots\right)$, the $x^2$ terms are:
$1(5x^2) + (ax)\left(-\frac{5}{8}x\right) + (3x^2)\left(\frac{1}{32}\right)$

3. Form an equation using the given coefficient of $x^2$ and solve for $a$.

Coefficient of $x^2 = 5 - \frac{5}{8}a + \frac{3}{32}$
Given that the coefficient is $\frac{13}{2}$:
$5 - \frac{5}{8}a + \frac{3}{32} = \frac{13}{2}$
$\frac{163}{32} - \frac{5}{8}a = \frac{208}{32}$
$-\frac{5}{8}a = \frac{45}{32}$
$a = -\frac{45}{32} \times \frac{8}{5} = -\frac{9}{4}$

1. Find the value of $x$ to substitute by equating the expression in the brackets to $0.47$.

$\frac{1}{2} - 2x = 0.47$
$0.5 - 2x = 0.47$
$2x = 0.03$
$x = 0.015$

2. Substitute $x = 0.015$ into the expansion from part (i).

From part (i), $\left(\frac{1}{2}-2x\right)^5 \approx \frac{1}{32} - \frac{5}{8}x + 5x^2 - 20x^3$

Substitute $x = 0.015$:
$(0.47)^5 \approx \frac{1}{32} - \frac{5}{8}(0.015) + 5(0.015)^2 - 20(0.015)^3$

3. Evaluate the expression and round to 5 decimal places.

$(0.47)^5 \approx 0.03125 - 0.009375 + 0.001125 - 0.0000675$
$(0.47)^5 \approx 0.0229325$
$(0.47)^5 \approx 0.02293 \text{ (to 5 d.p.)}$