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\begin{document}
  \begin{itemize}
  \item[1-]
  \begin{align*}
  &(a)\quad f(x)=-\dfrac{\pi}{4}+\dfrac{h}{2}+\sum _{k=1}^{\infty}\left\lbrace \frac{1}{\pi k^2}\left[ 1+(-1)^{k+1}\right] \cos kx\right. \\
  &\hspace{6.3cm}+\left. \frac{1}{\pi k}\left[h+(h+\pi)(-1)^{k+1} \right] \sin kx\right\rbrace \\
  &(c)\ \ f(x)=\sin x+\sum _{k=1}^{\infty}\frac{2(-1)^{k+1}}{k}\sin kx\\
  &(e)\ \ f(x)=\frac{\sinh \pi}{\pi}\left[ 1+\sum _{k=1}^{\infty}\frac{2(-1)^{k}}{1+k^2}(\cos kx-k\sin kx)\right] 
  \end{align*}
  \item[2-]
  \begin{align*}
  &\hspace{-3cm}(a)\ \ f(x)=\sum _{k=1}^{\infty}\dfrac{2}{k}\sin kx\\
  &\hspace{-3cm}(b)\ \ f(x)=\sum _{k=1}^{\infty}\left(\frac{2}{k\pi} \right) \left[1-2(-1)^k +\cos \frac{k\pi}{2} \right] \sin kx\\
  &\hspace{-3cm}(c)\ \ f(x)=\sum _{k=1}^{\infty}\left[2(-1)^{k+1}\dfrac{\pi}{k}+\frac{4}{\pi k^2}\left( (-1)^k-1\right)  \right] \sin kx\\
  &\hspace{-3cm}(d)\ \ f(x)=\sum _{k=2}^{\infty}\frac{2k}{\pi}\left[ \frac{1+(-1)^k}{k^2-1}\right] \sin kx
  \end{align*}
  \item[3-]
  \begin{align*}
  &\hspace{-2.5cm}(a)\ \ f(x)=\dfrac{3}{2}\pi +\sum _{k=1}^{\infty}\frac{2}{\pi k^2}\left[(-1)^k-1 \right]\cos kx\\
    &\hspace{-2.5cm}(b)\ \ f(x)=\dfrac{\pi}{2} +\sum _{k=1}^{\infty}\frac{2}{\pi k^2}\left[(-1)^k-1 \right]\cos kx\\
       &\hspace{-2.5cm}(c)\ \ f(x)=\dfrac{\pi ^2}{3} +\sum _{k=1}^{\infty}\frac{4(-1)^k}{k^2}\cos kx\\
  &\hspace{-2.5cm}(d)\ \ f(x)=\dfrac{2}{3\pi} +\sum _{k=1,2,4,\cdots}^{\infty}\frac{6}{\pi }\left[\frac{1+(-1)^k}{9-k^2} \right]\cos kx \qquad k\neq 3
  \end{align*}
  \item[4-]
  \begin{align*}
  &\hspace{-3.8cm}(b)\ \ f(x)=\sum _{k=1}^{\infty}\left(\frac{2}{k\pi} \right) \sin \frac{k\pi}{2}\cos \left(\frac{k\pi x}{6} \right) \\
   &\hspace{-3.8cm}(c)\ \ f(x)=\dfrac{2}{\pi}+\sum _{k=2}^{\infty}\left(\frac{2}{k\pi} \right)\left[\frac{1+(-1)^k}{1-k^2} \right]  \cos \left(\frac{k\pi x}{l} \right) \\
   &\hspace{-3.8cm}(f)\ \ f(x)=\sum _{k=1}^{\infty}\frac{k\pi}{1+k^2\pi ^2}(-1)^{k+1}\left( e-e^{-1}\right)\sin (k\pi x) 
  \end{align*}
  \item[5-]
  \begin{align*}
 &\hspace{-4.7cm}(a)\ \ f(x)=\sum _{k=-\infty}^{\infty}\dfrac{1}{\pi}\left( \frac{2+ik}{4+k^2}\right) (-1)^k\sinh 2\pi e^{ikx}\\
  &\hspace{-4.7cm}(b)\ \ f(x)=\sum _{k=-\infty}^{\infty}\frac{(-1)^k}{\pi \left( 1+k^2\right) } (-1)^k\sinh \pi e^{ikx}\\
   &\hspace{-4.7cm}(d)\ \ f(x)=\sum _{k=-\infty}^{\infty}(-1)^k\left( \frac{i}{k\pi}\right)  e^{ik\pi x}
   \end{align*}
   \item[6-]
   \[\hspace{-1.5cm}(a)\ \ f(x)=\dfrac{\pi}{8}\pi +\sum _{k=1}^{\infty}\left[ \frac{1}{2\pi k^2}\left\lbrace (-1)^k-1\right\rbrace \cos kx+\frac{(-1)^{k+1}}{2k}\sin kx \right]\]
   \item[7-]
   \[\hspace{-5.2cm}(a)\ \ f(x)=\frac{l^2}{3}+\sum _{k=1}^{\infty}4(-1)^k\left( \frac{1}{k\pi}\right)\cos \left(\frac{k\pi x}{l} \right)  \]
   \item[8-]
   \begin{align*}
   &\hspace{-4cm}(a)\ \ \sin ^2x=\sum _{k=1,3,4,\cdots}^{\infty}\frac{4(1-\cos k\pi}{k\pi \left( 4-k^2\right) }\sin kx\\
   &\hspace{-4cm}(b)\ \ \cos ^2x=\sum _{k=1,3,4,\cdots}^{\infty}\frac{2}{k\pi}\left( \frac{1-k^2}{ 4-k^2}\right) (1-\cos k\pi)\sin kx\\
    &\hspace{-4cm}(b)\ \ \sin x\cos x=\sum _{k=1,3,4,\cdots}^{\infty}\frac{2}{\pi}\left( \frac{1-\cos k\pi}{ 4-k^2}\right) \cos kx
    \end{align*}
    \item[9-]
    \begin{align*}
    &\hspace{-4cm}(a)\ \ \frac{x^2}{4}=\frac{\pi ^2}{12}-\sum _{k=1}^{\infty}\frac{(-1)^{k+1}}{k^2}\cos kx\\
    &\hspace{-4cm}(c)\ \ \int _{0}^{\infty}\ln \left( 2\cos \dfrac{x}{2}\right) \ud x=\sum _{k=1}^{\infty}(-1)^{k+1}\frac{\sin kx}{k^2}\\
    &\hspace{-4cm}(e)\ \ \dfrac{\pi}{2}-\dfrac{4}{\pi}\sum _{k=1}^{\infty}\frac{\cos (2k-1)x}{(2k-1)^2}=\left\lbrace  \begin{array}{ccc}
-x &  & -\pi <x<0 \\ 
 &  &  \\ 
x &  & 0<x<\pi
\end{array} \right.
\end{align*}
\item[10-]
\begin{align*}
&(a)\ \ f(x,y)=\frac{16}{\pi ^2}\sum _{m=1,3,\cdots}^{\infty}\sum _{n=1,3,\cdots}^{\infty}\left( \frac{1}{mn}\right)\sin mx\sin ny\\
&(c)\ \ f(x,y)=\frac{\pi ^4}{9}+\dfrac{1}{2}\sum _{m=1}^{\infty}\dfrac{8}{3}\pi ^2\frac{(-1)^m}{m^2}\cos mx+\dfrac{1}{2}\sum _{n=1}^{\infty}\dfrac{8}{3}\pi ^2\frac{(-1)^n}{n^2}\cos ny\\
&\hspace{2.5cm}+\sum _{m=1}^{\infty}\sum _{n=1}^{\infty} \frac{16(-1)^{m+n}}{m^2n^2}\cos mx\cos ny\\
&(e)\ \ f(x,y)=\sum _{m=1}^{\infty}\frac{2(-1)^{m+1}}{m}\sin mx \sin y\\
&(g)\ \ f(x,y)=\sum _{m=1}^{\infty}\sum _{n=1}^{\infty}d_{mn}\sin \left( \frac{m\pi x}{1}\right) \sin \left( \frac{n\pi y}{2}\right)\\
&\hspace{1cm}\begin{array}{ccccc}
d_{mn} & = & \frac{4}{1.2}\int _{0}^{2}\int _{0}^{1}xy\sin (m\pi x)\sin \left( \frac{n\pi y}{2}\right) \ud x\ud y&  &   \text{ که}   \\ 
&&&&\\
 & = & 2\int _{0}^{2}\left[\frac{\sin m\pi x}{m^2 \pi ^2}-\frac{x\cos m\pi x}{m\pi}\right] _{0}^{1}y\sin \left( \frac{n\pi y}{2}\right) \ud y &  &    \\
 &&&&\\ 
 & = &\hspace{-2cm} \frac{-2(-1)^m}{m\pi}\int _{0}^{2}y\sin \left( \frac{n\pi y}{2}\right) \ud y  =&\hspace{-2cm}\frac{-2(-1)^m}{m\pi}\left( \frac{-4(-1)^n}{n\pi}\right)  &    \\ 
 &&&&\\
 & =& \hspace{-4cm}\frac{8(-1)^{m+n}}{\pi ^2 mn} &  &    
\end{array} \\
&(h)\ \ f(x,y)=\left( \frac{16}{\pi ^2}\right)\sum _{m=1}^{\infty}\sum _{n=1}^{\infty}[(2m-1)(2n-1)]^{-1}\sin \left[ \frac{(2m-1)\pi x}{a}\right] \\
&\hspace{8.3cm}\times\sin \left[ \frac{(2n-1)\pi y}{a}\right]\\
 &(i)\ \ f(x,y)=\left( \frac{\sin 2}{\pi}\right) \sum _{m=1}^{\infty}\frac{(-1)^{m+1}}{m}\sin m\pi x\\
&\hspace{4cm}+\left( \frac{8\sin 2}{\pi}\right)\sum _{m=1}^{\infty}\sum _{n=1}^{\infty}\left[ \frac{(-1)^{m+n+1}}{m\left( 4-\pi ^2n^2\right) }\right] \sin (m\pi x)\cos \left( \frac{n\pi y}{2}\right) \\
&(j)\ \ f(x,y)=\dfrac{2}{3}\pi ^2 \sum _{m=1}^{\infty}\frac{(-1)^{m+1}}{m}\sin mx +\sum _{m=1}^{\infty}\sum _{n=1}^{\infty}\frac{8(-1)^{m+n+1}}{mn^2}\sin mx \sin ny\\
&(k)\ \ f(x,y)=\frac{\pi ^4}{9}+\left( \frac{4\pi ^2}{3}\right) \left[\sum _{m=1}^{\infty}\frac{(-1)^m}{m^2}\cos mx +\sum _{n=1}^{\infty} \frac{(-1)^n}{n^2}\cos ny\right] \\
&\hspace{6cm}+16\sum _{m=1}^{\infty}\sum _{n=1}^{\infty}\frac{(-1)^{m+n}}{m^2n^2}\cos mx\cos ny
\end{align*}
\item[20-]
\begin{flalign*}
&\hspace{-3cm}(a) \ \ b_{_{2n}}=0, \ \ b_{_{2n+1}}=\frac{8}{\pi(2n+1)^3}\\
&\hspace{-2.5cm}f(x)=x(\pi -x)=\dfrac{8}{\pi}\left(\frac{\sin x}{1^3}+ \frac{\sin 3x}{3^3} + \frac{\sin 5x}{5^3}+\cdots \right)\\
&\hspace{-3cm}(b)
\hspace{1cm} x=\dfrac{\pi}{2} \,  \text{ و } \, x=\dfrac{\pi}{4} \, \text{جهت یافتن مجموع سری ها قرار دهید}
\end{flalign*}
\item[21-]
\[ \hspace{-.8cm}(a) \ \ b_{n}=\dfrac{2}{\pi}\int _{0}^{\pi}f(x)\sin nx \ud x=\left( \frac{8}{n^2 \pi^{2}} \right) \sin \left( \frac{n\pi}{2}\right) ,\qquad n=0,1,2,\cdots\]
\[\hspace{-6.5cm} b_{_{2n}}=0,\quad b_n = \frac{8(-1)^{n}}{\pi ^2 (2n+1)^2}\]
\[\hspace{-7.2cm}(b) \hspace{3cm}x=\dfrac{\pi}{2}\ \ \text{قرار دهید}\]
\item[22-]
\begin{align*}
(a) \ \ f(x)& =\sum _{n=1}^{\infty} b_n \sin \left(\frac{n\pi x}{a}\right),\\
b_n &=\dfrac{2}{a}\int _0 ^a \sin \left(\frac{n\pi x}{a}\right) \ud x=\dfrac{2}{n\pi}(1-\cos n\pi) \\
&=\dfrac{2}{n\pi}\left[ 1-(-1)^n\right]=\left\lbrace 
\begin{array}{cc}
\frac{4}{n\pi} &\text{زوج} \; n \\ 
 &  \\ 
0 & \text{فرد} \; n
\end{array}\right. \\
f(x)&\sim \dfrac{4}{\pi}\left[ \sin \left(\frac{\pi x}{a}\right) +\dfrac{1}{3}\sin \left(\frac{3\pi x}{a}\right) + \dfrac{1}{5}\sin \left(\frac{5\pi x}{a}\right) + \cdots \right] \\
f(x)&\sim \dfrac{1}{2}a_0 + \sum _{n=1}^{\infty} a_n \cos \left(\frac{n\pi x}{a}\right),\qquad a_0 =2\\
a_n &=\frac{2}{n\pi}(\sin n\pi -0)=0,\qquad n\neq 0\\
1&= 1+0 \cdot \cos \left( \frac{n\pi}{a}\right)  +0\cdot \cos \left( \frac{2n\pi}{a}\right) + \cdots \\
(b)\ \ f(x)&=\sum _{n=1}^{\infty} (-1)^{n+1} \frac{2a}{n\pi} \sin \left(\frac{n\pi x}{a}\right)\\
f(x)& \sim \dfrac{1}{2}a_0 + \sum _{n=1}^{\infty} a_n \cos \left(\frac{n\pi x}{a}\right) \quad a_0 = \dfrac{2}{a}\int _0^a x \ud x =a \\
a_n &= \dfrac{2}{a}\int _0 ^a x\cos \left(\frac{n\pi x}{a}\right) \ud x =\frac{2a}{n^2 \pi ^2}\left( (-1)^n-1\right) \\
&=\left\lbrace \begin{array}{cc}
0 & \text{زوج} \; n \\ 
 &  \\ 
-\frac{4a}{n^2 \pi ^2} & \text{فرد} \; n
\end{array} \right. 
\end{align*}
\item[23-]
\begin{align*}
(a) \ \ a_0 &=\dfrac{1}{a}\int _{-a}^a x \ud x =0\\
a_n &=\dfrac{1}{a}\int _{-a}^a x\cos \left(\frac{n\pi x}{a}\right) \ud x =\left[ \frac{x}{n\pi}\sin \left(\frac{n\pi x}{a}\right) + \frac{a}{n^2 \pi ^2}\cos \left(\frac{n\pi x}{a}\right)\right] _{-a}^{a}\\
&=\frac{a}{n^2 \pi ^2}\left( \cos n\pi - \cos (-n\pi)\right) =0\\
b_n &= \dfrac{1}{a}\int _{-a}^a x \sin \left(\frac{n\pi x}{a}\right \ud x\\
&=\left[ \frac{-x}{n\pi}\cos \left(\frac{n\pi x}{a}\right) +\frac{a}{n^2 \pi ^2}\sin \left(\frac{n\pi x}{a}\right)\right] _{-a}^{a}\\
& =\frac{-a}{n\pi} \cos n\pi +\frac{-a}{n\pi}\cos (-n\pi)=(-1)^{n+1}\left( \frac{2a}{n\pi}\right) \\
(c)\ \ a_0 &= \frac{1}{2\pi}\int _0 ^{2\p1}f(x)\ud x  = \frac{1}{2\pi}\int _{\pi} ^{2\p1}\ud x =\dfrac{1}{2}\\
a_n &= \frac{1}{\pi}\int _0 ^{2\p1}f(x)\cos nx\ud x =\frac{1}{\pi}\int _{\pi} ^{2\p1}\cos nx\ud x =0 ,\quad n \; \text{برای هر}\\
b_n &= \frac{1}{\pi}\int _0 ^{2\p1}f(x)\sin nx\ud x =\frac{1}{\pi}\int _{\pi} ^{2\p1}\sin nx\ud x \\
&=\dfrac{1}{n\pi}\left[ -1+(-1)^n\right] =\left\lbrace \begin{array}{cc}
0 & \text{زوج} \; n \\ 
 &  \\ 
-\frac{2}{n\pi} & \text{فرد} \; n\right.
\end{array} \\
f(x)&= \dfrac{1}{2}-\dfrac{2}{\pi}\sum _{k=0}^{\infty}{\frac{\sin (2k+1)x}{2k+1}}
\end{align*}
 \end{itemize}
\end{document}