Use the binomial series to find the expansion of $$ \frac{1}{(2 + 5x)^3}, \\ |x| < \frac{2}{5} $$ in ascending powers of $x$, up to and including the term in $x^3$ - Edexcel - A-Level Maths Pure - Question 3 - 2016 - Paper 4

Start by rewriting the expression:

$$\frac{1}{(2 + 5x)^3} = \frac{1}{2^3} \cdot \frac{1}{(1 + \frac{5x}{2})^3} = \frac{1}{8} \cdot (1 + \frac{5x}{2})^{-3}$$

Now, apply the binomial series:

$$(1 + u)^{-3} = \sum_{k=0}^{\infty} \binom{-3}{k} u^k = \sum_{k=0}^{\infty} (-1)^k \frac{3(3 + 1)(3 + 2)\ldots (3 + k - 1)}{k!} u^k$$

Substituting $u = \frac{5x}{2}$ gives:

$$(1 + \frac{5x}{2})^{-3} = \sum_{k=0}^{\infty} (-1)^k \binom{3+k-1}{k} \left(\frac{5x}{2}\right)^k$$

We need to collect terms up to $x^3$:

• For $k=0$: inom{3}{0}(\frac{5}{2})^0 = 1
• For $k=1$: inom{4}{1}(\frac{5}{2})^1 = 20(\frac{5}{2}) = 50
• For $k=2$: inom{5}{2}(\frac{5}{2})^2 = 10(\frac{25}{4}) = \frac{250}{4} = 62.5
• For $k=3$: inom{6}{3}(\frac{5}{2})^3 = 20(\frac{125}{8}) = \frac{2500}{8} = 312.5 Hence, combining these terms, we have:

$$\frac{1}{8} (1 - 50x + 62.5x^2 - 312.5x^3)$$

Multiplying by rac{1}{8} gives:

$$\frac{1}{8} - \frac{50}{8}x + \frac{62.5}{8}x^2 - \frac{312.5}{8}x^3$$

Thus, the expansion up to $x^3$ is:

$$\frac{1}{8} - \frac{25}{4}x + \frac{15.625}{4}x^2 - \frac{39.0625}{4}x^3$$

With coefficients in their simplest form, we have: