Fourier Series Coefficients Question:

The fourier series coefficients is given as:

$$c_k= \begin{cases}
 1 \qquad &amp; k \ \text{ even} \\
 2 \qquad &amp; k \ \text{ odd} \\
\end{cases}$$

the period of the signal is $T=4$, what is signal $x(t)$?

Attempt:
when i am trying to find the signal by applying the general formula at the end i am getting exponential term now i am finding that hard to convert it into impulse again because i know answer of this signal will be a impulse but i am not getting any idea to convert it into impulse again? Please help with this?
You should use the synthesis equation of an impulse train with period $T$ (which is easy to derive):


  $$x(t)=\sum_{k=-\infty}^{\infty}\delta(t-kT)=\sum_{k=-\infty}^{\infty}\frac{1}{T}e^{jk\frac{2\pi}{T} t}\tag{1}$$
  That is: the Fourier coefficients for all terms is a constant ($\frac{1}{T}$). 


Now assume that there are two impulses with different amplitudes $a$ and $b$ per period, or $$x(t)=a\delta(t)+b\delta(t+2),\ -4&lt; t \le0$$ In such case we have 

$$\begin{align}
x(t)&amp;=\sum_{k=-\infty}^{\infty}a\cdot\delta(t-kT)+b\cdot\delta(t+\frac{T}{2}-kT)\\&amp;=\sum_{k=-\infty}^{\infty}a\cdot\delta(t-4k)+b\cdot\delta(t+2-4k)\\
&amp;=\sum_{k=-\infty}^{\infty}\frac{a}{4}e^{jk\frac{2\pi}{4} t}+\sum_{k=-\infty}^{\infty}\frac{b}{4}e^{jk\frac{2\pi}{4} (t+2)}\\
&amp;=\sum_{k=-\infty}^{\infty}\frac{1}{4}e^{jk\frac{\pi}{2} t}\left(a+be^{jk\pi}\right)\\
&amp;=\sum_{k=-\infty}^{\infty}\frac{1}{4}e^{jk\frac{\pi}{2} t}\left(a+b(-1)^k\right)\\
&amp;=\begin{cases}
\displaystyle\sum_{k=-\infty}^{\infty}\frac{a+b}{4}e^{jk\frac{\pi}{2} t},&amp; k\text{ even}\\[10pt]
\displaystyle\sum_{k=-\infty}^{\infty}\frac{a-b}{4}e^{jk\frac{\pi}{2} t},&amp; k\text{ odd}
\end{cases}
\end{align}$$ 
Now refering again to $(1)$ and comparing it with the question, we should have 
$$c_k=\begin{cases}
\frac{a+b}{4}=1\\[10pt]
\frac{a-b}{4}=2
\end{cases}$$