Q regarding finding sum of first 2002 terms

M:1 = Telescoping

I can see we can’t do anything after this. So , I use telescoping method w/ the help of user dxiv.

$\begin{aligned} a_{n+2} &=a_{n+1}-a_{n} \\ \therefore \quad a_{n+3} &=a_{n+2}-a_{n+1} \\ &=a_{n+1}-a_{n}-a_{n+1} \\ &=-a_{n} \end{aligned}$ $$ \begin{array}{ll}a_{n+4} & =a_{n+3}-a_{n+2} \\ \therefore \quad a_{n+5} & =a_{n+4}-a_{n+3} \\ \therefore \quad a_{n+4} & +a_{n+5}=a_{n+4}-a_{n+2} \\ \therefore \quad a_{n+5} & =-a_{n+2} \\ \therefore \quad a_{n}+a_{n+1}+a_{n+2}+a_{n+3}+a_{n+4}+a_{n+5} & \\ & =a_{n}+a_{n+1}+a_{n+2}-a_{n}+a_{n+3}-a_{n+2}-a_{n+2} \\ & = & a_{n+1}+a_{n+3}-a_{n+2}=0\end{array} $$ Let $S_{n}$ denotes the sum of first $n$ terms $$ \begin{aligned} S_{999} &=S_{6 \times 166+3}=S_{3} &(\because \text { every '6' consecutive }\\ S_{1003} &=S_{6 \times 167+1}=S_{1}^{\circ} &\text { terms has sum zero. }) \\ S_{2002} &=S_{6 \times 333+4}=S_{4} & \\ S_{2002} &=S_{4}=a_{1}+a_{2}+a_{3}+a_{4} \\ &=S_{3}-a_{1} \\ &=1003-(-999) \\ &=2002 \end{aligned} \quad\left(\because a_{4}=-a_{1} \text { since } a_{n+3}=-a_{n}\right) $$