Define x⋆y=√x2+3xy+y2−2x−2y+4xy+4. Compute ((⋯((2007⋆2006)⋆2005)⋆⋯)⋆1).
thanks a lot!
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−1n+4(an−1+n−1)2+3(an−1n+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.