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PROFESSOR LUCAS VISITS THE PUTNAM EXAMINATION Leon C* Woodson Department of Mathematics, Morgan State University, Baltimore, MD 21239-4001 (Submitted February 1996-Final Revision July 1996) The following problem appeared on the 1995 William Lowell Putnam Mathematical Competition: Evaluate 12207 - l 2207-2207-... and express the answer in the form (a + bjc)/d, where a, b, c, and id are integers. Readers of this journal might recognize that 2207 is the sixteenth Lucas number, Ll6. Therefore, a more general problem is to evaluate ^2n T In n n Let S denote this expression. Then S = L2n-(l/S ), lows that and therefore S2n-L2nSn +1 = 0. It fol- C" _ ^2n + V^2n ~ 4 A " 2 • Now, using the Binet formula for the Lucas numbers, i.e., Ln-an fi = (l + j5)/2, we have vn __ u — a2n +02n + ^(a2n +pnf-4 : — : : + J3n, a = (l + V5)/2, and __ a2n +02n + ^(a2n -j32n~f _ — 2 It follows that S = a = (3 + V5) / 2 (independent of n). ~ 2n — (X 2 2 This technique can be used to simplify a variety of expressions of this general form. A more natural solution to the original problem is to set T, say, equal to the expression and note, as above, that 7 1 6 -2207J 8 + 1 = 0. This can be written in the form Tl6+2T*+1 = 47 2 J 8 . Then, taking square roots (T is positive), we have r 8 + l = 477\ Repeating this gives T4 + l = 7T2 and T2 + l = 3T. Solving the latter equation yields the solution T = (3 + V5) 12. AMS Classification Numbers: 11A55, 11B39 1997] 341