Given the function: $$f(x) = \frac{3x^2 + 16}{(1 - 3x)(2 + x^2)}$$ (a) Find the values of A and C and show that B = 0 - Edexcel - A-Level Maths Pure - Question 7 - 2006 - Paper 7

Using the values found for A, B, and C, we can now write:

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

The series expansion for each term is:

• For the first term:

$\frac{3}{(1 - 3x)} = 3\sum_{n=0}^{\infty} (3x)^n = 3(1 + 3x + 9x^2 + 27x^3 + \ldots)$

Thus, the first term contributes:

$3 + 9x + 27x^2 + 81x^3$

• For the second term, since B = 0, it contributes nothing.

• For the third term, using the expansion for $\frac{4}{(2 + x^2)}$:

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

Combining the expansions, we get:

$f(x) = 3 + 9x + 27x^2 + 81x^3 + 2 - \frac{x^2}{1}$

The final simplified answer up to x^3 is:

$5 + 9x + (27 - 1)x^2 + 81x^3 = 5 + 9x + 26x^2 + 81x^3$.