Expand $$\frac{1}{(2-5x)^2}$$ in ascending powers of $x$, up to and including the term in $x^2$, giving each term as a simplified fraction - Edexcel - A-Level Maths Pure - Question 4 - 2012 - Paper 8

To expand $\frac{1}{(2-5x)^2}$ using the binomial series, we start with the general form:

$(1 + u)^n = \sum_{k=0}^{\infty} \binom{n}{k} u^k, \text{ where } |u| < 1$

For our case, let (u = -\frac{5x}{2}) and (n = -2).

We rewrite it as:

$(2-5x)^{-2} = 2^{-2} (1 - \frac{5x}{2})^{-2} = \frac{1}{4} \sum_{k=0}^{\infty} \binom{-2}{k} (-\frac{5x}{2})^k$

Calculating the first three terms:

• For (k=0): (\binom{-2}{0} (0) = 1) → (\frac{1}{4})
• For (k=1): (\binom{-2}{1} (-\frac{5x}{2}) = -2\cdot (-\frac{5x}{2}) = 5x) → (\frac{5x}{4})
• For (k=2): (\binom{-2}{2} (-\frac{5x}{2})^2 = 1\cdot \frac{25x^2}{4} = \frac{25x^2}{16}) → (\frac{25x^2}{64})

Thus, the expansion yields:

$\frac{1}{4} + \frac{5x}{4} + \frac{25x^2}{64}$.