+77 Define \(x \star y = \frac{\sqrt{x^2+3xy+y^2-2x-2y+4}}{xy+4}.\) Compute \(((\cdots ((2007 \star 2006) \star 2005) \star \cdots ) \star 1).\)
thanks a lot!
+160 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.
