$\displaystyle Let\\[2ex] c=-\pi{}\ and\ c+2l=\pi{}. \\[2ex] \displaystyle \therefore{}2l=\pi{}-c=\pi{}-\left(-\pi{}\right)=2\pi{} \\[2ex] \displaystyle \therefore{}l=\pi{}. \\[2ex] \begin{align*} \displaystyle Now,\\ a_0 &=\frac{1}{l}\int_c^{c+2l}f\left(x\right)dx \\[2ex] \displaystyle &={\frac{1}{\pi{}}\int_{-\pi{}}^{\pi{}}\left\vert{}x\right\vert{}dx \\[2ex]= \frac{2}{\pi{}}\int_0^{\pi{}}x\ dx\ \\[2ex]=\ \frac{2}{\pi{}}\left[\frac{{\ x}^2}{2}\right]}_0^{\pi{}} \\[2ex] \end{align*}$

$(Because \ \vert{}x\vert{} \ i s an \ even \ function \ and \vert{}x\vert{} = \ x \ in \ the \ interval \ 0 \ to \ \pi{} ) \\[2ex] \displaystyle =\frac{1}{\pi{}}\ [({\pi{})}^2-0]=\pi{} \\[2ex] \begin{align*} \therefore{}a_n&= \frac{1}{l} \int_c^{c+2l}f\left(x\right)\cos\frac{n\pi{}x}{l}dx\\[2ex] & =\frac{1}{\pi{}}\int_{-\pi{}}^{\pi{}}\vert{}x\vert{} \cos⁡(\frac{n\pi{}x}{l})dx\\[2ex] & =\frac{2}{\pi{}}\int_0^{\pi{}}x \cos\frac{n\pi{}x}{l}dx\\[2ex] &=\ \frac{2}{\pi{}}\int_0^{\pi{}}x\cos (nx)dx \\[2ex] \displaystyle \{since& \ x \cos {\left(n\frac{\pi{}x}{l}\right)}\ is\ an\ even\ function\ of\ x\}\ \\[2ex] \displaystyle &=\frac{2}{\pi{}}{\left[x.(\frac{sin\ nx}{n})-\frac{d\left(x\right)}{dx}\int{}\frac{sin\ nx}{n}dx\right]}_0^{\pi{}}\\[2ex]= &\frac{2}{\pi{}}{\left[x.(\frac{sin\ nx}{n})-1.\frac{-cos\ nx}{{\ n}^2}\right]}_0^{\pi{}} \\[2ex] \displaystyle &=\frac{2}{\pi{}}\left[\left(\pi{}.\frac{sin\ \pi{}n}{n}+\frac{cos\ n\pi{}}{{\ n}^2}\right)-\left(0+\frac{cos\ 0}{{\ n}^2}\right)\right]\ \\[2ex] \displaystyle \therefore{}a_n&=\frac{2}{\pi{}}\left[0+\frac{{\left(-1\right)}^n}{{\ n}^2}-0-\frac{1}{{\ n}^2}\right]\\[2ex] \displaystyle &=(2\frac{(-{1)}^n-1}{\pi{}n^2}) \\[2ex] Since, &\sin{n\pi{}}=0 ,\cos{n\pi{}} = {\left(-1\right)}^n\}\\[2ex] \displaystyle b_n&=\ \frac{1}{l}\int_c^{c+2l}f\left(x\right)sin\frac{n\pi{}x}{l}dx \\[2ex] &=\frac{1}{\pi{}}\int_{-\ \pi{}\ }^{\pi{}}\vert{}x\vert{}sin\frac{n\pi{}x}{l}dx \vert{}x\vert{} is an even function of x. \displaystyle \sin{n\frac{\pi{}x}{l}} \\[2ex] \end{align*}\\ $

$\vert{}x\vert{} is \ an \ even \ function \ of \ x. \displaystyle \sin{n\frac{\pi{}x}{l}} \\[2ex] is \ an odd \ function of \ x \ because \\[2ex] \displaystyle \sin{n\frac{\pi{}(-x)}{l}}=-\sin{n\frac{\pi{}x}{l}} \\[4ex] Hence, \vert{}x\vert{} \sin{n\frac{\pi{}x}{l}} \ will \ be \ an \ ODD \ function \ of \ x.\\[2ex] \displaystyle \therefore{}\ \int_{-\ \pi{}\ }^{\pi{}}\vert{}x\vert{}sin\frac{n\pi{}x}{l}dx=0\ \\[4ex] \displaystyle \therefore{}b_n=0\ \\[5ex] \displaystyle In\ fourier\ series, \\[2ex] \displaystyle \ \ f(x)=\ \frac{a_0}{2}+\sum_{n=1}^{\infty{}}a_n\ cos\frac{n\pi{}x}{l}+\sum_{n=1}^{\infty{}}b_n\ sin\frac{n\pi{}x}{l} \\[3ex] \displaystyle \therefore{}\vert{}x\vert{}=\frac{\pi{}}{2}+\sum_{n=1}^{\infty{}}2\frac{{\left(-1\right)}^n-\ 1}{\pi{}n^2}cos\frac{n\pi{}x}{\pi{}}+0\ \\[3ex] \displaystyle \therefore{}\vert{}x\vert{}=\frac{\pi{}}{2}-\frac{4}{\pi{}}\left[\frac{cos\ 1x}{1^2}+\frac{cos\ 3x}{3^2}+\frac{cos\ 5x}{5^2}+\ …\right] \\[5ex] \displaystyle Deduction\ :\ put\ x\ =\ \pi{}\ , \\[4ex] \displaystyle \cos{1\pi{}=\cos{3\pi{}=\cos{5\pi{}=…=-1}}} \\[2ex] \displaystyle \therefore{}\pi{}=\frac{\pi{}}{2}+\frac{4}{\pi{}}\left[\frac{1}{1^2}+\frac{1}{3^2}+\frac{1}{5^2}+…\right] \\[2ex] \begin{align*} \displaystyle \therefore{}\left[\frac{1}{1^2}+\frac{1}{3^2}+\frac{1}{5^2}+…\right]&=\ \frac{\pi{}}{\ 4}\left(\pi{}-\ \frac{\pi{}}{2}\right)\\[2ex]& =\frac{\pi{}}{4}\Bigg(\frac{\pi{}}{2}\Bigg) \\[2ex]&=\ \frac{{\pi{}}^2}{8} \\[2ex] \end{align*}$