Let, $f(x)=a_0+\sum_{n=1}^{\infty}\ a_n\ cos\ nx+\sum_{n=1}^{\infty}b_n\ cosnx$

$\therefore a_0=\dfrac{1}{2\pi }\int_0^{2\pi }\ f(x)\ dx=\dfrac{1}{2\pi }\int_0^{2 \pi }\dfrac{(3x^2-6x\pi +2\pi ^2)}{12}dx$

$a_0=\dfrac{1}{24\pi }\Big[x^3-3x^2\pi +2\pi ^2x\Big]_0^{2\pi }=\dfrac{1}{24\pi }\Big[8\pi ^3-12\pi ^3+4\pi ^3\Big]=0$

$a_n=\dfrac{1}{\pi }\int_0^{2\pi }\ f(x)cos\ nx\ dx=\dfrac{1}{\pi }\int_0^{2\pi }\Bigg(\dfrac{3x^2-6x\pi +2\pi ^2}{12}\Bigg)cos\ nx\ dx$

$a_n=\dfrac{1}{12\pi }\Bigg[3\Bigg[x^2\Bigg(\dfrac{sin\ nx}{n}\Bigg)-(2x)\Bigg(-\dfrac{cos\ nx}{n^2}\Bigg)+(2) \Bigg(-\dfrac{sin \ nx}{n^3}\Bigg)\Bigg]-6\pi \Bigg[x \Bigg(\dfrac{sin \ nx}{n}\Bigg)-(1)\Bigg(-\dfrac{cos\ nx}{n^2} \Bigg)\Bigg]+2\pi ^2\Bigg(\dfrac{sin\ nx}{n}\Bigg)\Bigg]_0^{2\pi }$

$a_n=\dfrac{1}{12\pi }\Bigg[3\Bigg(0+\dfrac{4\pi }{n^2}+0\Bigg)-6\pi \Bigg(0+\dfrac{1}{n^2}\Bigg)+2\pi ^2(0)-\Bigg(0-6\pi \Bigg(\dfrac{1}{n^2}\Bigg)\Bigg)\Bigg]=\dfrac{1}{n^2}$

$b_n=\dfrac{1}{\pi }\int_0^{2\pi }\ f(x)\ sin\ nx\ dx=\dfrac{1}{\pi }\int_0^{2\pi }\Bigg(\dfrac{3x^2-6x\pi +2\pi ^2}{12}\Bigg)\ sin\ nx\ dx\\ b_n=\dfrac{1}{12\pi }\Bigg[3\Bigg(x^2 \Bigg(-\dfrac{cos\ nx}{n}\Bigg)-(2x)\Bigg(-\dfrac{sin nx}{n^2}\Bigg)+2(1) \Bigg(\dfrac{cos\ nx}{n^3}\Bigg) \Bigg)-6\pi \Bigg[x\Bigg(-\dfrac{cos\ nx}{n}\Bigg)-(1)\Bigg(-\dfrac{sin\ nx}{n^2}\Bigg)\Bigg]+2\pi ^2\Bigg(-\dfrac{cos\ nx}{n}\Bigg)\Bigg]_0^{2\pi }$

$b_n=\dfrac{1}{12\pi }\Bigg[3\Bigg(-\dfrac{4\pi ^2}{n}+\dfrac{2}{n^3}\Bigg)-6\pi \Bigg(-\dfrac{2\pi }{n} \Bigg)-\dfrac{2\pi ^2}{n}-\Bigg(3\Bigg(\dfrac{2}{n^3}\Bigg)-\dfrac{2\pi ^2}{n}\Bigg)\Bigg]=0$

$\therefore \ f(x)=\sum a_n\ cos\ nx=\sum\dfrac{1}{n^2}\ cos\ nx$

$\therefore \ \dfrac{3x^2-6x\pi +2\pi ^2}{12}=\dfrac{1}{1^2}\ cos\ x+\dfrac{1}{2^2}\ cos\ 2x+\dfrac{1}{3^2}\ cos 3x+......$

Putting x = 0 we get, 

$\dfrac{2\pi ^2}{12}=\dfrac{1}{1^2}+\dfrac{1}{2^2}+\dfrac{1}{3^2}+.......\\ \therefore \dfrac{\pi ^2}{6}=\dfrac{1}{1^2}+\dfrac{1}{2^2}+\dfrac{1}{3^2}+.....$