f(x) = (3 + 2x)^{-3}, |x| < \frac{1}{2} - Edexcel - A-Level Maths Pure - Question 3 - 2007 - Paper 7

Now, we calculate the first few terms of the expansion:

$f(x) = \frac{1}{27} \left(\binom{-3}{0}(\frac{2x}{3})^0 + \binom{-3}{1}(\frac{2x}{3})^1 + \binom{-3}{2}(\frac{2x}{3})^2 + \binom{-3}{3}(\frac{2x}{3})^3 + \ldots \right)$

Calculating the coefficients:

• For $k=0$: $\binom{-3}{0} = 1$
• For $k=1$: $\binom{-3}{1} = -3$
• For $k=2$: $\binom{-3}{2} = \frac{(-3)(-4)}{2!} = 6$
• For $k=3$: $\binom{-3}{3} = \frac{(-3)(-4)(-5)}{3!} = -10$

Substituting these values back into the expansion gives:

$f(x) = \frac{1}{27} \left( 1 - 3 \cdot \frac{2x}{3} + 6 \cdot \left(\frac{2x}{3}\right)^2 - 10 \cdot \left(\frac{2x}{3}\right)^3 \right)$

Calculating the terms gives:

• Constant term: $\frac{1}{27}$
• Linear term: $-2x$
• Quadratic term: $\frac{8x^2}{27}$
• Cubic term: $-\frac{80x^3}{27}$

Combining everything, the final result will be:

$f(x) = \frac{1}{27} - 2x + \frac{8}{27}x^2 - \frac{80}{27}x^3 + \ldots$

Each coefficient has been simplified as fractional values.