PROBLEMS from volume 31 (1993) issue #3 of The Fibonacci Quarterly.

Copyright ◯^{c} 1993 by The Fibonacci Association.

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ELEMENTARY PROBLEMS AND SOLUTIONS Edited by Stanley Rabinowitz

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BASIC FORMULAS

The Fibonacci numbers F_{n} and the Lucas numbers L_{n} satisfy

F_{n+2} = F_{n+1} + F_{n}, F_{0} = 0, F_{1} = 1;
L_{n+2} = L_{n+1} + L_{n}, L_{0} = 2, L_{1} = 1.

Also, α = (1 + \sqrt{5})/2, β = (1 − \sqrt{5})/2, F_{n} = (α^{n} − β^{n})/\sqrt{5}, and L_{n} = α^{n} + β^{n}.

PROBLEMS PROPOSED IN THIS ISSUE

\underline{B–742} Proposed by Curtis Cooper and Robert E. Kennedy, Central Missouri State University, Warrensburg, MO

Pell numbers are defined by P_{0} = 0, P_{1} = 1, and P_{n+1} = 2P_{n} + P_{n−1}, for n ≥ 1. Show that

P_{23} = 2^{11} ∏_{j=1}^{11} (3 + cos \frac{2jπ}{23} ) .

\underline{B-743} Proposed by Richard André-Jeannin, Longwy, France Find the modulus and the argument of the complex numbers

u = \frac{β + i\sqrt{α + 2}}{2}  and  v = \frac{α + i\sqrt{β + 2}}{2}.

\underline{B–744} Proposed by Herta T. Freitag, Roanoke, VA

Let n and k be even positive integers. Prove that L_{2n} + L_{4n} + L_{6n} + ⋯ + L_{2kn} is divisible by L_{n}.

\underline{B–745} Proposed by Richard André-Jeannin, Longwy, France Show that

∑_{n=1}^{∞} \frac{1}{F_{2n}} = 1 + ∑_{n=1}^{∞} \frac{1}{F_{2n−1}F_{2n}F_{2n+1}} .

\underline{B–746} Proposed by Seung-Jin Bang, Albany, CA Solve the recurrence equation a_{n+1} = 4a_{n}^{3} + 3a_{n}, n ≥ 0, with initial condition a_{0} = 1/2.

\underline{B–747} Proposed by Piero Filipponi, Fond. U. Bordoni, Rome, Italy Let

S_{1} = ∑_{n=2}^{∞} \frac{1}{(−1)^{n}L_{2n−1} − 1}  and  S_{2} = ∑_{n=2}^{∞} \frac{1}{(−1)^{n}L_{2n−1} + 1}.

Prove that S_{1}/S_{2} = \sqrt{5}.