Solution:

$\displaystyle Let \ f(x)=a_0+\sum_{n=1}^{\infty{}}\left(a_n\cos{\frac{n\pi{}x}{L}}+b_n\sin{\frac{n\pi{}x}{L}}\right) $

$Here \ 2l=2, \ \therefore \ l=1 \\ f\left(x\right)=a_0+\sum_{n=1}^{\infty{}}\left(a_n\cos{n\pi{}x}+b_n\sin{n\pi{}x}\right) ...(1)$

$\displaystyle a_0=\ \frac{1}{2l}\int_0^{2l}f\left(x\right)\ dx \\[2ex] \displaystyle =\frac{1}{2}\left[\int_0^1\pi{}x\ dx\ +\int_1^2\pi{}(2-x)\ dx\ \right] \\[2ex] \displaystyle =\frac{1}{2}\left[\pi{}{\left\{\frac{x^2}{2}\right\}}_0^1+\pi{}{\left\{2x-\frac{x^2}{2}\right\}}_1^2\right] \\[2ex] \displaystyle =\frac{\ \pi{}}{2}\left[\left\{\frac{1}{2}\right\}+\left\{4-2-2+\frac{1}{2}\right\}\right] \\[2ex] =\dfrac{\pi{}}{2}$

$a_n=\ \ \frac{1}{l}\int_0^{2l}f\left(x\right)\ cos\ n\pi{}x\ dx \\[2ex] =\int_0^1\pi{}x\ cosn\pi{}x\ dx\ +\int_1^2\pi{}(2-x)\ cosn\pi{}x\ dx\\[2ex] =\pi{}\left[{\left\{x\dfrac{\sin{n\pi{}x}}{n\pi{}}-\left(1\right)\dfrac{-cos{n\pi{}x}}{n^2{\pi{}}^2}\right\}}_0^1+\ {\left\{(2-x)\dfrac{\sin{n\pi{}x}}{n\pi{}}-\left(-1\right)\dfrac{-cos{n\pi{}x}}{n^2{\pi{}}^2}\right\}}_1^2\right] \\[2ex] =\pi{}\left[\left\{\dfrac{\cos{n\pi{}}}{n^2{\pi{}}^2}-\dfrac{1}{n^2{\pi{}}^2}\right\}+\ \left\{-\dfrac{1}{n^2{\pi{}}^2}+\dfrac{\cos{n\pi{}}}{n^2{\pi{}}^2}\right\}\right] \\[2ex] =\pi{}\left[\dfrac{2cos{n\pi{}}}{n^2{\pi{}}^2}-\dfrac{2}{n^2{\pi{}}^2}\right] \\[2ex] =\dfrac{2\pi{}}{n^2{\pi{}}^2}[cos n\pi{} - 1] $

$=\dfrac{2}{n^2\pi{}}[{(-1)}^n- 1] \\[2ex] \left\{\begin{array}{l}0\ \ \ \ \ \ \ \ \ \ \ \ \ \ if\ n\ is\ even \\ -\dfrac{4}{n^2\pi{}}\ \ \ \ \ if\ n\ is\ odd\end{array}\right.$

$b_n=\ \dfrac{1}{l}\int_0^{2l}f\left(x\right)\ sinn\pi{}x\ dx$

$=\int_0^1\pi{}x\ sinn\pi{}x\ dx\ +\int_1^2\pi{}(2-x)\ sinn\pi{}x\ dx$

$=\pi{}\left[{\left\{x\dfrac{-cos{n\pi{}x}}{n\pi{}}-\left(1\right)\dfrac{-sin{n\pi{}x}}{n^2{\pi{}}^2}\right\}}_0^1+\ {\left\{(2-x)\dfrac{-cos{n\pi{}x}}{n\pi{}}-\left(-1\right)\dfrac{-sin{n\pi{}x}}{n^2{\pi{}}^2}\right\}}_1^2\right]$

$=\pi{}\left[\left\{\dfrac{-cos{n\pi{}}}{n\pi{}}\right\}+\ \left\{\dfrac{\cos{n\pi{}}}{n\pi{}}\right\}\right] \\[2ex] =0$

Substituting these values in (1)...

$\displaystyle f\left(x\right)=\frac{\ \pi{}}{2}+\sum_{n=1}^{\infty{}}\left(\frac{2}{n^2\pi{}}[{(-1)}^n\ -\ 1]\cos{n\pi{}x}\right)$

$\displaystyle f\left(x\right)=\frac{\ \pi{}}{2}-\ \frac{4}{\pi{}}\left[\frac{\cos{\pi{}x}}{1^2}+\frac{\cos{\pi{}x}}{3^2}+\frac{\cos{\pi{}x}}{5^2}+…\right]$