Let $f(x) = \frac{3x^2 + 16}{(1-3x)(2+x)^3} + \frac{A}{(1-3x)} + \frac{B}{(2+x)^2} + \frac{C}{(2+x)^3}, \quad |x| < 1$ - Edexcel - A-Level Maths Pure - Question 6 - 2006 - Paper 7

Using the values from part (a), the function simplifies as:

$f(x) = \frac{3x^2 + 16}{(1-3x)(2+x)^3} + \frac{3}{(1-3x)} + 0 + \frac{4}{(2+x)^3}$

We can expand the individual components.

For the term ( \frac{3}{(1-3x)} ):

Using the geometric series: $\frac{3}{(1-3x)} = 3 \sum_{n=0}^{\infty} (3x)^n = 3 + 9x + 27x^2 + \dots$

For ( \frac{4}{(2+x)^3} ), we can apply the binomial expansion:

$\frac{4}{(2+x)^3} = 4 \sum_{n=0}^{\infty} {\binom{-3}{n} (2)^{-3-n} x^n}$

Calculating the first few terms:

• When n=0, it's 1,
• When n=1, it's ( -\frac{12}{8} \cdot 1 = 4 \ )

By combining all the series expansions together, we can finalize:

$f(x) = 3 + 9x + (27 + 4) x^2 + O(x^3) = 3 + 9x + 31x^2 + O(x^3)$