Keeping the Fibonacci rule of adding the latest two to get the next but starting from 2 and 1 (in this order) instead of 0 and 1 for the (ordinary Fibonacci numbers) one defines the series of Lucas Numbers:
L(n) = L(n-1) + L(n-2) for n>1
L(0) = 2
L(1) = 1
Here are some more values of L(n) together with the Fibonacci numbers
for comparison:
F(n): 0 1 1 2 3 5 8 13 21 34 55 ...
L(n): 2 1 3 4 7 11 18 29 47 76 123 ...
With the above information in background prove the following formulas:
a. F(n-1)+ F(n+1) = L(n)
b. L(n-1) + L(n+1)= 5*F(n)
c. F(n)* L(n) = F(2n)
exploring...
L(0) = F(0) + 2
L(1) = F(1)
L(2) = L(0) + L(1)
= F(0) + F(1) + 2
= F(2) + 2
L(3) = L(1) + L(2)
= F(1) + F(2) + 2
= F(3) + 2
L(4) = L(2) + L(3)
= F(2) + 2 + F(3) + 2
= F(4) + 4
L(5) = L(3) + L(4)
= F(3) + 2 + F(4) + 4
= F(5) + 6
etc.
So this becomes our fundamental formula:
L(n) = F(n) + 2 F(n-1) (for n>0)
a) Show F(n-1)+ F(n+1) = L(n)
F(n-1) + F(n+1)
= F(n-1) + F(n-1) + F(n)
= 2 F(n-1) + F(n)
= L(n)
Q.E.D.
b) Show L(n-1) + L(n+1)= 5*F(n)
L(n-1) + L(n+1)
= F(n-1) + 2 F(n-2) + F(n+1) + 2 F(n)
= F(n) + F(n-2) + F(n+1) + 2 F(n)
= F(n) + F(n-2) + F(n) + F(n-1) + 2 F(n)
= 3 F(n) + 2 F(n)
= 5 F(n)
Q.E.D.
Edited on September 10, 2018, 10:39 am
a and b soln.s
