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Define xy=x2+3xy+y22x2y+4xy+4. Compute ((((20072006)2005))1).

 

thanks a lot!

 Oct 12, 2024
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Observe that x2+3xy+y2−2x−2y+4​=(x+y−1)2+3(xy+4)​.

 

Thus, x⋆y=xy+4(x+y−1)2+3(xy+4)​​. Let an​=(((⋯((2007⋆2006)⋆2005)⋆⋯)⋆n).

 

Then we can write the recurrence relation an​=an−1​n+4(an−1​+n−1)2+3(an−1​n+4)​​.

 

Note that a1​=1. We can compute a2​,a3​,… using this recurrence relation. A

 

fter some computations, we find that a_2007 = 2008/2007.

 Oct 12, 2024

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