Solution:

Let, $f(x)=\frac{1}{4}(\pi-x)^{2}=\frac{a_{0}}{2}+\sum_{n=1}^{\infty} a_{n} \cos n x+\sum_{n=1}^{\infty} b_{n}\sin nx,\\$ $By\ Euler's formula, we\ have$ $$ \begin{aligned} a_{0} &=\frac{1}{\pi} \int_{0}^{2 \pi} f(x) dx \end{aligned} $$

$$=\frac{1}{\pi} \int_{0}^{2 \pi} \frac{1}{4}(\pi-x)^{2} d x$$

$$=\frac{1}{4 \pi}\left[\frac{(\pi-x)^{3}}{-3}\right]_{0}^{2 \pi}$$

$$=-\frac{1}{12 \pi}\left[-\pi^{3}-\pi^{3}\right]=\frac{\pi^{2}}{6}$$

$$ \begin{aligned} a_{n} &=\frac{1}{\pi} \int_{0}^{2 \pi} f(x) \cos n x d x=\frac{1}{\pi} \int_{0}^{2 \pi} \frac{1}{4}(\pi-x)^{2} …