find sum of first 2002 terms

Question

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 $$\sum_{n=1}^{2002} a_n $$

Answer 1

$$a_3+a_2+a_1 = (a_2 - a_1) + a_2 + a_1 = 2 a_2\\ a_4+a_3+a_2+a_1 = (a_3 - a_2) + 2a_2 = a_3 + a_2\\ a_5+a_4+\dots a_1 = (a_4 -a_3) + a_3 + a_2 = a_4+a_2$$

In general, the formula $$\sum_{i=1}^na_i = a_{n-1} + a_2$$ can be proven by induction: the case for $n=3$ is already proven, now $$\sum_{i=1}^{n+1}a_i = a_{n+1} + \sum_{i=1}^na_i = (a_{n} - a_{n-1}) + a_{n-1} + a_2 = a_n + a_2$$ proves the induction step.

Using this, I can see that $a_{998} + a_2 = 1003$ and that $a_{1002} + a_2 = -999$. From the two equations for the sums, I also get that $$a_{1003} + a_{1002} + a_{1001} + a_{1000} = - 2002$$ or $$(a_{1002} - a_{1001}) + a_{1002} + a_{1001} + a_{1000} = -2002\\ 2a_{1002} + a_{1000} = -2002\\ 2(a_{1001} - a_{1000}) + a_{1000} = -2002\\ 2a_{1001} - a_{1000} = -2002$$

This is as far as I got so until now...

EDIT: I believe the sequence $a_i$ is periodic:

$$a_3=a_2-a_1\\ a_4=-a_1\\ a_5 = -a_2\\ a_6=-a_2+a_1\\ a_7=a_1\\ a_8=a_2\\ $$ Since $a_{7}$ and $a_{8}$ equal $a_1$ and $a_2$, the same pattern of $6$ figures (from $a_1$ to $a_{6}$ will just repeat itself for all values of $n$. This means that $$a_{1002} = a_{996} = a_{990} = \dots = a_6 = -a_2 + a_1\\ a_{998} = a_{992} = a_{984} = \dots a_2$$

Plugging them into the equations $a_{998} + a_2 = 1003$ and that $a_{1002} + a_2 = -999$ gives me

$$a_2 + a_2 = 1003\\ a_2 = \frac{1003}{2}$$ and $$-a_2 + a_1+a_2 = -999\\ a_1 = -999$$

Meaning that $$\sum_{n=1}^{2002} a_n = a_{2001} + a_2 = a_{3} + a_2 = a_2 - a_1 + a_2 = 2a_2-a_1 = 1003 + 999 = 2002$$

Answer 2

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*}