1. The first two terms of the sequence are $a_{1}=1$ and $a_{2}=3$. From the second term onwards, it holds that if we subtract the term before it from a given term, we get the term after it. Find the sum of the first 2005 terms of the sequence.
IV /1. For $i \geq 2$, it holds that $a_{i+1}=a_{i}-a_{i-1}$. Thus, $a_{1}+a_{2}+a_{3}+\cdots+a_{2005}=a_{1}+a_{2}+\left(a_{2}-\right.$ $\left.a_{1}\right)+\left(a_{3}-a_{2}\right)+\left(a_{4}-a_{3}\right)+\cdots+\left(a_{2003}-a_{2002}\right)+\left(a_{2004}-a_{2003}\right)=a_{2}+a_{2004}$. If we write out the first few terms of the sequence, we observe that the sequence is periodic with a period of 6: $1,3,2,-1,-3$, $-2,1,3 \ldots$ Since $2004=6 \cdot 334+0$, it follows that $a_{2004}=a_{6}=-2$. The sum of the first 2005 terms of the sequence is $a_{1}+a_{2}+a_{3}+\cdots+a_{2005}=a_{2}+a_{2004}=3-2=1$.

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