$\displaystyle C=0\ \ \ \ \ \ c+2l=2\ \ \ i.e\ 0+2l=2\ \ \ \ \ \therefore{}l=1 \\[2ex] \displaystyle Now,\\[2ex] \displaystyle a_0=\frac{1}{l}\int_c^{c+2l}f\left(x\right)dx \\[2ex] \displaystyle ={\frac{1}{1}\int_0^24-{\ x}^2dx\\[2ex] \displaystyle =1\left[{\ 4x-\frac{x^3}{3}x}^3\ \right]}_0^2 \\[2ex] \displaystyle =\left[\ 4\left(2\right)-\frac{2^3}{3}\ \right]–[0-0] \\[2ex] \displaystyle =8-\frac{8\ }{3}=\ \frac{16}{3} \\[2ex] $

$\displaystyle \therefore{}a_n=\ \frac{1}{l}\int_c^{c+2l}f\left(x\right)\cos\frac{n\pi{}x}{l}dx \\[2ex] \displaystyle =\frac{1}{1}\int_0^24-{\ x}^2\cos{(n\pi{}x})dx \\[2ex] \displaystyle =1(4-{\ x}^2)\frac{\sin\left(n\pi{}x\right)}{n\pi{}}-\left(0-2x\right)-\frac{\cos{\left(n\pi{}x\right)}}{{\ n}^2{\ \pi{}}^{2\ }} \\[2ex] \displaystyle =[{\ 0-\frac{4\bullet{}1}{{\ n}^2{\ \pi{}}^{2\ }}}^{\ }+0]-\left[0-0+0\right]\\[2ex] \displaystyle =-\frac{4}{{\ n}^2{\ \pi{}}^{2\ }} \\[2ex] $

$\displaystyle \therefore{}b_n=\ \frac{1}{l}\int_c^{c+2l}f\left(x\right)\sin\frac{n\pi{}x}{l}dx \\[2ex] \displaystyle =\frac{1}{1}\int_0^24-{\ x}^2\sin{\left(\frac{n\pi{}x}{l}\right)}dx \\[2ex] \displaystyle =1{\left[(4-{\ x}^2)-\left(\frac{\cos\ n\pi{}x}{n\pi{}}\right)-\left(-2x\right)\bullet{}\frac{\sin{\left(n\pi{}x\right)}}{{\ n}^2{\ \pi{}}^{2\ }}+(-2)\frac{\cos{\left(n\pi{}x\right)}}{{\ n}^3{\ \pi{}}^{3\ }}\right]}_0^2 \\[2ex] \displaystyle =\left [0-0-2\bullet{}\frac{1}{{\ n}^3{\ \pi{}}^{3\ }}\right]-\left[4.-\frac{1}{n\pi{}}-0-\frac{2\bullet{}1}{{\ n}^3{\ \pi{}}^{3\ }}\right]\\[2ex] \displaystyle =\frac{4}{n\pi{}}\ \\[2ex] $

$\displaystyle In\ fourier\ series, \\[2ex] \displaystyle f(x)=\ \frac{a_0}{2}+\sum_{n=1}^{\infty{}}a_n\cos\frac{n\pi{}x}{l}+\sum_{n=1}^{\infty{}}b_n\sin\frac{n\pi{}x}{l} \\[2ex] \displaystyle \therefore{}4-x^2=\frac{8}{3}-\sum_{n=1}^{\infty{}}-\frac{4}{{\ n}^2{\ \pi{}}^{2\ }}\cos\frac{n\pi{}x}{1}+\sum_{n=1}^{\infty{}}\frac{4}{n\pi{}}\ \sin\frac{n\pi{}x}{1} \\[2ex] $

$\displaystyle =\frac{8}{3}-\frac{4}{{\ \pi{}}^{2\ }}\left[\frac{1}{{\ 1}^2}\cos\ 1\pi{}x+\frac{1}{{\ 2}^2}\cos\ 2\pi{}x+…\right]+\frac{4}{\pi{}}\left[\frac{1}{1}\sin{\left(\pi{}x\right)}+\frac{1}{2}\sin{\left(2\pi{}x\right)}+…\right]\ \ \ \ \ ....(1) \\[4ex] Deduction: \\[2ex] $

$Put\ x=0\ in\ (1) \\[2ex] \displaystyle 4-0\ =\ \frac{8}{3}-\frac{4}{{\ \pi{}}^{2\ }}\left[\frac{1}{1^2}\left(1\right)\frac{1}{{\ 2}^2}\left(1\right)+…\right]+\frac{4}{\pi{}}\left[0+0+…\right]\ \ \ \ \\[4ex] \displaystyle 4-\frac{8}{3}=\ -\frac{4}{{\ \pi{}}^{2\ }}\left[\frac{1}{{\ 1}^2}\left(1\right)\frac{1}{{\ 2}^2}\left(1\right)+…\right] \\[4ex] \displaystyle \therefore{}\frac{4}{3}\times{}\frac{{\ \pi{}}^{2\ }}{-4}=\left[\frac{1}{{\ 1}^2}\left(1\right)\frac{1}{{\ 2}^2}\left(1\right)+…\right] \\[2ex] \displaystyle \therefore{}-\frac{{\ \pi{}}^{2\ }}{3}=\frac{1}{{\ 1}^2}+\frac{1}{{\ 2}^2}+\frac{1}{{\ 3}^2}+.…(2)\\[4ex] $

$Put\ x=2\ in\ (1)\\[2ex] \displaystyle {4-\ 2}^2\ =\ \frac{8}{3}-\frac{4}{{\ \pi{}}^{2\ }}\left[\frac{1}{1^2}\left(1\right)+\frac{1}{{\ 2}^2}\left(1\right)+…\right]+\frac{4}{\pi{}}\left[0+0+…\right]\ \ \ \ \ \\[2ex] \displaystyle 0-\frac{8}{3}=\ -\frac{4}{{\ \pi{}}^{2\ }}\left[\frac{1}{{\ 1}^2}+\frac{1}{{\ 2}^2}+\frac{1}{{\ 3}^2}…\right] \\[2ex] \displaystyle \therefore{}\frac{-8}{3}\times{}\frac{{\ \pi{}}^{2\ }}{-4}=\left[\frac{1}{{\ 1}^2}+\frac{1}{{\ 2}^2}+\frac{1}{{\ 3}^2}…\right] \\[2ex] $

$\displaystyle \therefore{}\frac{{\ 2\pi{}}^{2\ }}{3}=\frac{1}{{\ 1}^2}+\frac{1}{{\ 2}^2}+\frac{1}{{\ 3}^2}…(3)\\[3ex]Adding\ (2)\ and\ (3),\\[2ex] \displaystyle \therefore{}\frac{{\ 2\pi{}}^{2\ }}{3}-\frac{{\ \pi{}}^{2\ }}{3}=\frac{2}{{\ 1}^2}+\frac{2}{{\ 2}^2}+\frac{2}{{\ 3}^2}+… \\[2ex] \displaystyle \therefore{}\frac{{\ \pi{}}^{2\ }}{3}=2\left [\frac{1}{{\ 1}^2}+\frac{1}{{\ 2}^2}+\frac{1}{{\ 3}^2}+…\right] \\[2ex] \displaystyle \therefore{}\frac{{\ \pi{}}^{2\ }}{6}=\frac{1}{{\ 1}^2}+\frac{1}{{\ 2}^2}+\frac{1}{{\ 3}^2}+… \\[2ex] $