2021 WSMO Speed Round Problems/Problem 3

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Problem

Let $a@b=\frac{a^2-b^2}{a+b}$. Find the value of $1@(2@(\dots(2020@2021)\dots)$.

Solution

Note that $a^2-b^2=(a-b)(a+b).$ This means that $a@b=a-b.$ Now, $a@(b@c)=a-(b-c)=a-b+c$ and $a@(b@(c@d))=a-(b-(c-d))=a-b+c-d.$ Using this pattern, we find that \[1@(2@(\dots(2020@2021)\dots)=1-2+3-4+\ldots-2020+2021=\]\[(1-2)+(3-4)+\ldots+(2019-2020)+2021=-1010+2021=\boxed{1011}.\]

~pinkpig