+0  
 
0
216
1
avatar+644 

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
 #1
avatar+93038 
+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

8 Online Users

avatar

New Privacy Policy

We use cookies to personalise content and advertisements and to analyse access to our website. Furthermore, our partners for online advertising receive information about your use of our website.
For more information: our cookie policy and privacy policy.