solution:

Let ,$x-x^{2}=\frac{a_{0}}{2}+\sum_{n=1}^{\infty} a_{n} \cos n x+\sum_{n=1}^{\infty} b_{n} \sin n x$ By Euler's formulae, we have,

$ a_{0}=\frac{1}{\pi} \int_{-\pi}^{\pi}\left(x-x^{2}\right) d x=\frac{1}{\pi}\left[\frac{x^{2}}{2}-\frac{x^{3}}{3}\right]_{-\pi}^{\pi}\\ $

$$ =\frac{1}{\pi}\left[\left(\frac{\pi^{2}}{2}-\frac{\pi^{3}}{3}\right)-\left(\frac{\pi^{2}}{2}+\frac{\pi^{3}}{3}\right)\right]=-\frac{2 \pi^{2}}{3}\\ $$

$$ \begin{aligned} a_{n} &=\frac{1}{\pi} \int_{-\pi}^{\pi}\left(x-x^{2}\right) \cos n x d x \\\\ &=\frac{1}{\pi}\left[\left(x-x^{2}\right) \frac{\sin n x}{n}-(1-2 x)\left(-\frac{\cos n x}{n^{2}}\right)+(-2)\left(-\frac{\sin n x}{n^{3}}\right)\right]_{-\pi}^{n} \\\\ &=\frac{1}{\pi}\left[(1-2 \pi) \frac{\cos …