find sum of first 2002 terms if $\left \{ a_n \right \}$ is sequence of Real Numbers for $n \ge 1$ such that
\begin{equation}
a_{n+2}=a_{n+1}-a_n \tag{1}  
\end{equation}
\begin{equation}
\sum_{n=1}^{999} a_n=1003 \tag{2}
\end{equation}
\begin{equation}
\sum_{n=1}^{1003}a_n=-999 \tag{3}
\end{equation}
Then Find the value of $$\sum_{n=1}^{2002} a_n $$
 A: Anaother way to solve the problem is to use the characteristic equation. The characteristic equation is 
$$ r^2=r-1 $$
whose solution is $r_1=e^{\frac{\pi}{3}i}, r_2=e^{-\frac{\pi}{3}i}$. So $a_n$ can be expressed as
$$ a_n=C_1e^{\frac{n\pi}{3}i}+C_2e^{-\frac{n\pi}{3}i} $$
where $C_1, C_2$ are constants which will be determined later. Hence
\begin{eqnarray*}
S_n&=&\sum_{k=1}^na_n=C_1\sum_{k=1}^ne^{\frac{k\pi}{3}i}+C_2\sum_{k=1}^ne^{-\frac{k\pi}{3}i}\\
&=&C_1\left(\frac{1-e^{\frac{(n+1)\pi}{3}i}}{1-e^{\frac{\pi}{3}i}}-1\right)+C_2\left(\frac{1-e^{-\frac{(n+1)\pi}{3}i}}{1-e^{-\frac{\pi}{3}i}}-1\right).
\end{eqnarray*}
Easy calculation shows that
$$ a_{999}=(-1+\sqrt{3}i)C_1+(-1-\sqrt{3}i)C_2, a_{1003}=\frac{1+\sqrt{3}i}{2}C_1+\frac{1-\sqrt{3}i}{2}C_1. $$
Solving
$$ a_{999}=1003, a_{1003}=-999 $$
gives
$$ C_1=-\frac{3001}{4}+\frac{995i}{4\sqrt{3}}, C_2=-\frac{3001}{4}-\frac{995i}{4\sqrt{3}} $$
and hence
\begin{eqnarray*}
a_{2002}&=&C_1e^{\frac{2002\pi}{3}i}+C_2e^{-\frac{2002\pi}{3}i}\\
&=&-C_1e^{\frac{\pi}{3}i}-C_2e^{-\frac{\pi}{3}i} \\
&=&Re[-(-\frac{3001}{4}+\frac{995i}{4\sqrt{3}})\frac{1+\sqrt{3}i}{2}]\\
&=&2002.
\end{eqnarray*}
