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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.)

waffles  Feb 14, 2018
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+1

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

 

 

cool cool cool

CPhill  Feb 14, 2018

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