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

L_0 = 2

L_1 = 1

L_n = L_(n - 1) + L_(n - 2)

then what is the 8th Lucas number? (Note: We seek a numerical answer.)

Guest Oct 16, 2020

#2**+1 **

Since you have already stated the first 2 Lucas numbers, then it is simply a matter of ADDING them up !.

L_0 = 2, L_1 = 1. L_2? It is simply the sum of L_0 + L_1 =2 + 1 = 3. L_3 =L_2 + L_1 =3 + 1 =4. L_4 =4 + 3 =7.

L_5 =7 + 4 =11. L_6 =11 + 7 =18. L_7 =18 + 11 =29. **L_8 = 29 + 18 =47......and so on.**

They go like this: 2 , 1 , 3 , 4 , 7, 11, 18 ,29, 47.....and so on. Each Lucas number is simply the SUM of the previous two. So:47 =29 + 18. 29 = 18 + 11. 18 =11 + 7. 11 = 7 + 4. 7 =4 + 3. 4 =3 + 1. And 3 = 1 + 2. Got it?

Guest Oct 17, 2020