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}\\ $

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

$$ \begin{aligned} a_{n} &=\frac{1}{\pi} \int_{-\pi}^{\pi}\left(x-x^{2}\right) \cos n x d x \\\\ &=\frac{1}{\pi}\left[\left(x-x^{2}\right) \frac{\sin n x}{n}-(1-2 x)\left(-\frac{\cos n x}{n^{2}}\right)+(-2)\left(-\frac{\sin n x}{n^{3}}\right)\right]_{-\pi}^{n} \\\\ &=\frac{1}{\pi}\left[(1-2 \pi) \frac{\cos n \pi}{n^{2}}-(1+2 \pi) \frac{\cos n \pi}{n^{2}}\right]=\frac{1}{\pi}\left(-4 \pi \cdot \frac{\cos n \pi}{n^{2}}\right) \\\\ &=-4 \frac{(-1)^{n}}{n^{2}} \\\\ b_{n} &=\frac{1}{\pi} \int_{-\pi}^{\pi}\left(x-x^{2}\right) \sin n x d x \end{aligned} $$

$ \left[\because \cos n \pi=(-1)^{n}\right]\\ $

$$ \begin{aligned} &=\frac{1}{\pi}\left[\left(x-x^{2}\right)\left(-\frac{\cos n x}{n}\right)-(1-2 x)\left(-\frac{\sin n x}{n^{2}}\right)+(-2)\left(\frac{\cos n x}{n^{3}}\right)\right]_{-\pi}^{n} \\\\ &=\frac{1}{\pi}\left[\left(\pi^{2}-\pi\right) \frac{\cos n \pi}{n}-2 \frac{\cos n \pi}{n^{3}}+\left(-\pi-\pi^{2}\right) \frac{\cos n \pi}{n}+2 \frac{\cos n \pi}{n^{3}}\right] \\\\ &=\frac{1}{\pi}\left[-2 \pi \cdot \frac{\cos n \pi}{n}\right]=-2 \frac{(-1)^{n}}{n} \\\\ \therefore \quad x-x^{2} &=-\frac{\pi^{2}}{3}-4 \sum_{n=1}^{-} \frac{(-1)^{n}}{n^{2}} \cos n x-2 \sum_{n=1}^{-} \frac{(-1)^{n}}{n} \sin n x \\\\ &=-\frac{\pi^{2}}{3}-4\left[-\frac{\cos x}{1^{2}}+\frac{\cos 2 x}{2^{2}}-\frac{\cos 3 x}{3^{2}}+\ldots \ldots\right]\\\\ =&-\frac{\pi^{2}}{3}+4\left[\frac{\cos x}{1^{2}}-\frac{\cos 2 x}{2^{2}}+\frac{\cos 3 x}{3^{2}}-\ldots\ldots\right]+2\left[\frac{\sin x}{1}-\frac{\sin 2 x}{2}+\frac{\sin 3 x}{3}-\ldots \ldots\right] \end{aligned}\\\\ $$

Putting, $x=0$,

$ we \ get\ 0=-\frac{\pi^{2}}{3}+4\left(\frac{1}{1^{2}}-\frac{1}{2^{2}}+\frac{1}{3^{2}}-\frac{1}{4^{2}}+\ldots \ldots\right) \\ $

$$ \Rightarrow \quad \frac{1}{1^{2}}-\frac{1}{2^{2}}+\frac{1}{3^{2}}-\frac{1}{4^{2}}+\ldots \ldots=\frac{\pi^{2}}{12} \text {. }\\ $$