How to find the Fourier Coefficient to this discrete time signal? I'm having trouble finding the fourier coefficients to this discrete time signal:

My approach:
I start by first finding the discrete period, $N$, and discrete frequency, $\Omega_0$
From the problem, we can easily identify that $N = 10$ and so $\Omega_{0} = \frac{2\pi}{10} = \frac{\pi}{5}$
$$x[n] = \sum_{k = 0}^{9}F[K]e^{j\left( \frac{\pi}{5}\right)kn} $$
where $F[K]$ are the Fourier coefficients we want to find.
$$F[K] = \frac{1}{N}\sum_{n = 0}^{9}x[n]e^{-j\left( \frac{\pi}{5}\right)kn}$$
$$F[K] = \frac{1}{10}\left( \sum_{n = 0}^{4}\frac{7}{2}e^{-j\left( \frac{\pi}{5}\right)kn} + \sum_{n = 4}^{9}\frac{-7}{2}e^{-j\left( \frac{\pi}{5}\right)kn} 
\right)$$
$$F[K] = \frac{7}{20}\left( \sum_{n = 0}^{4}e^{-j\left( \frac{\pi}{5}\right)kn} - \sum_{n = 4}^{9}e^{-j\left( \frac{\pi}{5}\right)kn} 
\right)$$
$$F[K] = \frac{7}{20} \left( 1 + \cos(\frac{\pi}{5}k) + j\sin(\frac{\pi}{5}k) + \cos(\frac{2\pi}{5}k) + j\sin(2\frac{\pi}{5}k) + \cos(\frac{3\pi}{5}k) + j\sin(3\frac{\pi}{5}k) + \cos(\frac{4\pi}{5}k) + j\sin(\frac{4\pi}{5}k) - \cos(\pi k) - j\sin(\pi k) - \cos(\frac{6\pi}{5}k) - j\sin(\frac{6\pi}{5}k) - \cos(\frac{7\pi}{5}k) - j\sin(\frac{7\pi}{5}k) - \cos(\frac{8\pi}{5}k) - j\sin(\frac{8\pi}{5}k) - \cos(\frac{9\pi}{5}k) - j\sin(\frac{9\pi}{5}k)
\right)$$
This looks like such a mess to reduce... or is there something elegant that I'm missing?
According to the solutions
$
X[K]=
\begin{cases}
 \frac{7}{10} + \frac{7j \pi}{5}\left( \sin(\frac{\pi}{5}) + \sin( \frac{2\pi}{5}) \right) , &\text{if}\, k = - 1\\
 \frac{7}{10} + \frac{7j \pi}{5}\left( \sin(\frac{\pi}{5}) + \sin( \frac{2\pi}{5}) \right) , &\text{if}\ k = + 1\\
 \frac{7}{10} - \frac{7j \pi}{5}\left( \sin(\frac{2 \pi}{5}) - \sin( \frac{\pi}{5}) \right) , &\text{if}\ k = + 3\\
 \frac{7}{10} + \frac{7j \pi}{5}\left( \sin(\frac{2 \pi}{5}) - \sin( \frac{\pi}{5}) \right) , &\text{if}\ k = - 3\\
      0 &\text{otherwise}
\end{cases}
$
HOW DO YOU FIGURE THAT OUT!!? Is there some super clever technique I'm not employing? Or is all my math wrong?
 A: Your second sum should index from $n=5$.  Then here is a small simplification:
\begin{align}
F[k] &= \frac{7}{20} \Big( \sum_{n=0}^4 e^{-ink\pi/5}-\sum_{n=5}^9e^{-ink\pi/5} \Big) \\
&=\frac{7}{20} \Big( \sum_{n=0}^4 e^{-ink\pi/5}-\sum_{n=0}^4   e^{-5ik \pi/5} e^{-ink\pi/5} \Big) \\
&= \frac{7(1 - (-1)^k)}{20} \sum_{n=0}^4 e^{-ink\pi/5} \\
&=\left\{ \begin{array} ~0 & \text{when } k \text{ is even} \\
\frac{7}{10} \sum_{n=0}^4 e^{-ink\pi/5} & \text{otherwise} \\
\end{array}\right .
\end{align}
Thus for even $k$, $F[k] = 0$.  Otherwise, pair the terms in the sum using the fact that $k$ is odd,
\begin{align}
e^{-ik\pi/5}+e^{-i4k\pi/5}&=e^{-ik\pi/5}-e^{ik\pi/5}=-2i\sin(k\pi/5), \quad\text{ pairing } n = 1 \text{ and } n=4\\
e^{-i2k\pi/5}+e^{-i3k\pi/5}&=e^{-2ik\pi/5}-e^{i2k\pi/5}=-2i\sin(2k\pi/5), \quad\text{ pairing } n = 2 \text{ and } n = 3 .
\end{align}
So taking the real and imaginary, the $\cos$ terms in the sum cancel for all values of $k$ and we are left only with two $\sin$ terms in the imaginary part,
\begin{align}
F[k]=\left\{ 
\begin{array}
~0 & \text{when } k \text{ is even} \\
~\frac{7}{10} -\frac{7}{5}i \Big(\sin\big(\frac{1}{5}k\pi\big)+\sin\big(\frac{2}{5}k\pi\big) \Big)&\text{when } k \text{ is odd},
\end{array} \right.
\end{align}
which is close to the answer you wanted;  the stated answer has a $\pi$ factor in the imaginary part which I think must be an error, and we differ in some signs.
PS.  I have used $i$ instead of $j$ out of habit.
