(x^{n+1}
+ x^{(n+1)}) = (x^{n} + x^{n})(x + x^{1}) –
(x^{n1} + x^{(n1)}),
so writing f_{n} = x^{n} + x^{n} gives: f_{n+1}
= f_{n}*f_{1}  f_{n1} (1)
x^{2} + x + 1 = 0 => x(x + x^{1})
= x and, since x = 0 is
not a solution, we can divide by x to obtain
f_{1} = 1,
then (1) becomes: f_{n+1} = (f_{n}
+ f_{n1}).
From the definition, f_{0} = 2, so this Fibonaccilike
recurrence equation can be used to give
f_{0}, f_{1}, …….. = 2, 1, 1, 2, 1, 1, 2, 1, 1, 2……
and the required Sum_{1 to 9}(f_{n}^{2}) is:
1 + 1 + 4 + 1 + 1 + 4 + 1 + 1 + 4 = 18_{
}

Posted by Harry
on 20150115 08:20:20 