If x² + x +1 =0, then find the value of:
(x+1/x)² + (x² + x^-2)² + ...+ (x⁹ + x^-9)²

(xⁿ⁺¹ + x^-(n+1)) = (xⁿ + x^-n)(x + x^-1) – (x^n-1 + x^-(n-1)),

so writing f_n = xⁿ + x^-n  gives:      f_n+1 = f_n*f₁ - f_n-1     (1)

x² + x + 1 = 0 =>  x(x + x^-1) = -x and, since x = 0 is

not a solution, we can divide by x to obtain  f₁ = -1,

then (1) becomes:   f_n+1 = -(f_n + f_n-1).

From the definition, f₀ = 2, so this Fibonacci-like

recurrence equation can be used to give

f₀, f₁, …….. = 2, -1, -1, 2, -1, -1, 2, -1, -1, 2……

and the required Sum_{1 to 9}(f_n²) is:

1 + 1 + 4 + 1 + 1 + 4 + 1 + 1 + 4 = 18

Posted by Harry on 2015-01-15 08:20:20