+647 The Lucas numbers are defined in the same way, but with different starting values. Let L_0 be the zeroth Lucas number and L_1 be the first. If
\(\begin{align*} L_0 &= 2 \\ L_1 &= 1 \\ L_n &= L_{n - 1} + L_{n - 2} \; \text{for}\; n \geq 2 \end{align*}\)
then what is the tenth Lucas number? (Note: We seek a numerical answer.)
+128474 The nth Lucas number can be derived as follows
L(n) = Phi^n + (-phi)^n
Where
Phi = (1 + √5) / 2 ) and -phi = -2 / ( 1 + √5)
So
L(10) = [ 1 + [ ( (1 + √5) / 2 )^10 + ( -2 / ( 1 + √5) )^10 ] = 123
Also....if you know the Fibonacci Series....
L(n) = F(n - 1) + F(n + 1)
So
L(10) = F(9) + F(11) = 34 + 89 = 123
