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

Using the binomial series expansion for ((1 + u)^{-n}), which is:

[ (1 + u)^{-n} = \sum_{k=0}^{\infty} \binom{-n}{k} u^k ] we can substitute (u = \frac{5x}{2}) and (n = 3):

[ (2 + 5x)^{-3} = 2^{-3} \left(1 + \frac{5x}{2}\right)^{-3} = \frac{1}{8}\left(1 + \frac{5x}{2}\right)^{-3} ]

Expanding these terms gives:

[ = 1 - \frac{15x}{2} + \frac{75x^2}{4} - \frac{125x^3}{8} ]

Now, multiply the entire expansion by (\frac{1}{8}):

[ \frac{1}{8}\left(1 - \frac{15x}{2} + \frac{75x^2}{4} - \frac{125x^3}{8}\right) ] This results in:

[ = \frac{1}{8} - \frac{15x}{16} + \frac{75x^2}{32} - \frac{125x^3}{64} ]

Thus, the expansion of (\frac{1}{(2 + 5x)^3}) in ascending powers of (x) is:

[ \frac{1}{8} - \frac{15}{16}x + \frac{75}{32}x^2 - \frac{125}{64}x^3 ] with coefficients in their simplest form.