*$Answer:***  f(x) = 2x-1 Fourier series for f(x)=$a_0 +\sum a_ncos \dfrac{\pi n x}{l} +\sum b_n sin\dfrac{\pi n x}{l}.

here, 2l=3 ,l=\dfrac{3}{2} $ $a_0=\dfrac{1}{2l} \int_0^{2l} f(x).dx=\dfrac{1}{3} \int_0^{3} (2x-1).dx$ $=\dfrac{1}{3}[x^2-x]_0^3 = \dfrac{1}{3}.(9-3)\

=2$ $a_n = \dfrac{2}{3} .\int_o^3 (2x-1).cos\dfrac{n\pi x.3}{2 }.dx$ $\dfrac{2}{3}[(2x-1).sin( \dfrac{n\pi x2}{3}).\dfrac{3}{2\pi n}+ 2cos( \dfrac{n\pi x2}{3}).\dfrac{9}{4\pi^2 n^2}]_0^3\

\dfrac{2}{3} (0+\dfrac{9}{2\pi^2 n^2}-0-\dfrac{9}{2\pi^2 n^2 })\

=0$   $bn=\dfrac{1}{l} \int_0^l f(x) .sin\dfrac{\pi n x}{l}.dx\ \dfrac{2}{3} \int_0^3 (2x-1) .sin\dfrac{2\pi n x}{3}.dx$ $[(2x-1)(-cos\dfrac{2\pi nx }{3}.\dfrac{3}{2\pi n} +2.sin(\dfrac{2\pi nx } {3}).\dfrac{9}{4\pi^2 n^2}]_0^3$ $b_n=\dfrac{2}{3}[5.-(cos2\pi n.\dfrac{3}{2\pi n}) + 2.(sin2\pi n.\dfrac{9}{4\pi^2 n^2}).]\

\dfrac{2}{3} (5(-1)\dfrac{3}{2\pi n}1)+0)\

=-\dfrac{4}{\pi n}$ $f(x) =a_0 +\displaystyle\sum_{n=1}^\infty a_n cos\dfrac{2\pi nx}{3}

• \displaystyle\sum_{n=1}^\infty b_n sin\dfrac{2\pi nx}{3}\ \therefore f(x) =2 -\dfrac{4}{\pi} \displaystyle\sum_{n=1}^\infty \dfrac{1}{n} sin\dfrac {2\pi nx}{3}

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