Given the elementary signal $x(t)= 2\Pi(t/2) + 2 \Pi((t-5)/2)$, where $\Pi(t)$ is the rectangular pulse function.
The pulse train is built as $y(t)=\sum_{k=-\infty}^\infty x(t-10k)$, which can be rewritten as:
$$y(t)=2\sum_{k=-\infty}^\infty\left[\Pi\left(\frac{t-10k}{2}\right) + \Pi\left(\frac{t-10k-5}{2}\right)\right]$$
As you pointed out, the pulse train has fundamental period $T_0=5$. This basically means that the pulse train can be further simplified into:
$$y(t)=\sum_{k=-\infty}^\infty2\Pi\left(\frac{t-5k}{2}\right)$$
We will, therefore, use the elementary pulse centered at $t=0$ and with amplitude $2$, $f(t)=2\Pi\left(\frac{t}{2}\right)$ to perform our Fourier series expansion.
The Fourier series can be written as:
\begin{align}
f(t)&=c_{0}+\sum_{n=-\infty}^{-1}\left[c_{n}e^{j\cdot n\omega_0 t}\right]+\sum_{n=1}^{\infty}\left[c_{n} e^{j n \omega_0 t}\right]\\
c_{0}&=\frac{1}{T_0}\int_{T_0}f(t)dt\quad\text{(average value)}\\
c_{n}&=\frac{1}{T_0}\int_{T_0}f(t)e^{-jn \omega_0 t}dt
\end{align}
Where $\omega_0$ is the fundamental angular frequency of your signal: $\omega_0=2\pi f_0=\frac{2\pi}{T_0}=\frac{2\pi}{5}$
Let us first calculate the average value:
\begin{align}
c_{0}&=\frac{1}{5}\int_{-2.5}^{2.5}2\Pi\left(\frac{t}{2}\right)dt\\
&=\frac{2}{5}\int_{-1}^{1}dt\\
&=\frac{4}{5}
\end{align}
Now for the coefficients:
\begin{align}
c_{n}&=\frac{1}{5}\int_{-2.5}^{2.5}2\Pi\left(\frac{t}{2}\right)e^{-jn \omega_0 t}dt\\
&=\frac{2}{5}\int_{-1}^{1}e^{-jn \omega_0 t}dt\\
&=\frac{2}{5}\frac{e^{jn \omega_0}-e^{-jn \omega_0}}{jn\omega_0}\\
&=\frac{2}{n\pi} \sin\left(\frac{2n\pi}{5}\right)
\end{align}
Then you should just plug the coefficients back into the Fourier series representation and there you have it.
Hope this helps
Best Answer
Rephrasing your question:
Solution:
Let Fourier transform of $x(t)=e^{\frac{-2\pi}{|T_0|}|t-nT_0|}$ be $X(f)$ and that of $x^\prime(t)=e^{\frac{-2\pi}{|T_0|}|t|}$ be $X^\prime(f)$.
Time-shifting property of Fourier transform \begin{align*} y(t\pm t_0)&\xrightarrow{\text{FT}} Y(f)e^{\pm j2\pi ft_0}\\ \text{if}\quad y(t)&\xrightarrow{\text{FT}}Y(f) \end{align*} So, $$e^{\frac{-2\pi}{|T_0|}|t-nT_0|}\xrightarrow{\text{FT}} X^\prime(f)e^{-j2\pi fnT_0}\tag{1}$$ You know that $$e^{-a|t|}\xrightarrow{\text{FT}} \frac{2a}{a^2+(2\pi f)^2}$$ So, $$e^{\frac{-2\pi}{|T_0|}|t|}\xrightarrow{\text{FT}}\frac{\frac{4\pi}{|T_0|}}{\left(\frac{2\pi}{|T_0|}\right)^2+(2\pi f)^2}\equiv \frac1{\pi |T_0|}\left(\frac1{\frac1{(T_0)^2}+f^2}\right)\tag{2}$$ From equation $(1)$ and $(2)$, we get $$e^{\frac{-2\pi}{|T_0|}|t-nT_0|}\xrightarrow{\text{FT}} \frac1{\pi |T_0|}\left(\frac1{\frac1{(T_0)^2}+f^2}\right)e^{-j2\pi fnT_0}\tag{3}$$
On summating equation $(3)$ over all integers $n$, we get $$\sum_{n=-\infty}^{+\infty}e^{\frac{-2\pi}{|T_0|}|t-nT_0|}\xrightarrow{\text{FT}} \frac1{\pi |T_0|}\left(\frac1{\frac1{(T_0)^2}+f^2}\right)\sum_{n=-\infty}^{+\infty}e^{-j2\pi fnT_0}=\frac{f_0}{\pi}\left(\frac1{f_0^2+f^2}\right)\sum_{n=-\infty}^{+\infty}e^{+j2\pi fnT_0}\equiv \frac{1}{\pi f_0}\left(\frac1{1+\left(\frac{f}{f_0}\right)^2}\right)\sum_{n=-\infty}^{+\infty}e^{+j2\pi fnT_0}\quad\quad(\because n\in (-\infty,+\infty)\text{ and }T_0f_0=1) \tag{4}$$
We know that the Fourier series of an impulse train is $$\sum_{n=-\infty}^{+\infty}\delta(t-nT)=\frac1{T}\sum_{n=-\infty}^{+\infty}e^{j2\pi nt/T}\tag{5}$$ Finding an impulse train that could divert the RHS of equation $(4)$ towards the answer by replacing $t\rightarrow f$ and $T\rightarrow f_0$: $$\sum_{n=-\infty}^{+\infty}\delta(f-nf_0)=\frac1{f_0}\sum_{n=-\infty}^{+\infty}e^{j2\pi nf/f_0}\tag{6}$$ Replacing $k\rightarrow f/f_0$ and substituting equation $(6)$ in equation $(4)$, we get $$\sum_{n=-\infty}^{+\infty}e^{\frac{-2\pi}{|T_0|}|t-nT_0|}\xrightarrow{\text{FT}} \frac{1}{\pi}\left(\frac1{1+\left(\frac{f}{f_0}\right)^2}\right)\sum_{n=-\infty}^{+\infty}\delta(f-nf_0)=\frac{1}{\pi}\left(\frac1{1+k^2}\right)\delta(f-kf_0)\quad\quad(\because \delta(f-nf_0)=0\ \forall\ n \neq k)$$ (Hence proved :))