Final answer:
The Maclaurin series for f(x) = ln(1-5x) is -5x - \frac{{25x^2}}{2} - \frac{{125x^3}}{3} - \frac{{625x^4}}{4} - \frac{{3125x^5}}{5} - ..., and the interval of validity is -\frac{1}{5} < x < \frac{1}{5} or -0.2 < x < 0.2.
Step-by-step explanation:
The Maclaurin series for f(x) = ln(1-5x) can be found by expanding the natural logarithm function using its power series representation. The power series representation of ln(1-5x) is given by:
f(x) = -5x - \frac{{25x^2}}{2} - \frac{{125x^3}}{3} - \frac{{625x^4}}{4} - \frac{{3125x^5}}{5} - ...
The interval of validity for this series is the range of values of x for which the series converges. In this case, the series converges for -\frac{1}{5} < x < \frac{1}{5} or -0.2 < x < 0.2.