A) Find the binomial expansion of sqrt(1-8x), |x| < (1)/(8), in ascending powers of x up to and including the term in x3, simplifying each term. b) Show that, when x = (1)/(100), the exact value of sq

here are the solutions to question 2.

2. a) Find the binomial expansion of $\sqrt{1-8x}$, $|x| < \frac{1}{8}$, in ascending powers of $x$ up to and including the term in $x^3$, simplifying each term.

Step 1: Rewrite the expression in the form $(1+y)^n$.

$$\sqrt{1-8x} = (1-8x)^{\frac{1}{2}}$$

Here, $n = \frac{1}{2}$ and $y = -8x$.

Step 2: Apply the binomial expansion formula $(1+y)^n = 1 + ny + \frac{n(n-1)}{2!}y^2 + \frac{n(n-1)(n-2)}{3!}y^3 + \dots$.

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

Step 3: Simplify each term.

$$= 1 + \left(\frac{1}{2}\right)(-8x) + \frac{\frac{1}{2}\left(-\frac{1}{2}\right)}{2}(64x^2) + \frac{\frac{1}{2}\left(-\frac{1}{2}\right)\left(-\frac{3}{2}\right)}{6}(-512x^3) + \dots$$ $$= 1 - 4x + \frac{-\frac{1}{4}}{2}(64x^2) + \frac{\frac{3}{8}}{6}(-512x^3) + \dots$$ $$= 1 - 4x - \frac{1}{8}(64x^2) + \frac{3}{48}(-512x^3) + \dots$$ $$= 1 - 4x - 8x^2 - 32x^3 + \dots$$

The binomial expansion is $\boxed{1 - 4x - 8x^2 - 32x^3}$.