Let un= 2 cosh(Fn) then un+1un+2= eFn+1+ e−Fn+1
eFn+2+ e−Fn+2

Problem 11978

(American Mathematical Monthly, Vol.124, May 2017) Proposed by H. Ohtsuka (Japan).

Let Fn be the n-th Fibonacci number. Find

∞

X

n=0

(−1)ⁿ

cosh(Fⁿ) cosh(Fn+3).

Solution proposed by Roberto Tauraso, Dipartimento di Matematica, Universit`a di Roma “Tor Vergata”, via della Ricerca Scientifica, 00133 Roma, Italy.

Solution. Let uⁿ= 2 cosh(Fⁿ) then

u_n+1u_n+2= e^Fn+1+ e^−Fn+1
e^Fn+2+ e^−Fn+2

= e^Fn+1+Fn+2+ e^Fn+2−Fn+1+ e^−Fn+2+Fn+1+ e^{−Fn+1−Fn+2}

= e^Fn+3+ e^Fn+ e^−Fn+ e^−Fn+3

= uⁿ+ un+3. Hence

N

X

n=0

(−1)ⁿ

cosh(Fⁿ) cosh(Fn+3) = 4

N

X

n=0

(−1)ⁿ u_nu_n+3 = 4

N

X

n=0

(−1)ⁿ(un+ un+3) u_nu_n+1u_n+2u_n+3

= 4

N

X

n=0

 (−1)ⁿ

u_n+1u_n+2u_n+3− (−1)ⁿ⁻¹ u_nu_n+1u_n+2



= 4

 (−1)^N

u_{N +1}u_{N +2}u_{N +3} −

−1 u₀u₁u₂

N →∞

→ 4

u₀u₁u₂ = 1 2 cosh²(1)

.