Question

The Lucas numbers are defined recursively as follows:

$L_1=1,L_2=3,$and $L_k=L_{k-2}+L_{k-1}$for $k\ge 3$.

What are$L_3,L_4,L_5,$and$L_6$?

Short answer

The required values are$L_3=4,L_4=7,L_5=11,L_6=18$.

Step-by-step solution

Step 1. Given information

$L_1=1,L_2=3$.

$L_k=L_{k-2}+L_{k-1}$where$k\ge 3$.

Step 2. Substitute k=3 and L1=1,L2=3 in the given sequence.

$$\begin{gathered} L_3=L_{3-2}+L_{3-1} \\ =L_1+L_2 \\ =1+3 \\ =4 \end{gathered}$$

Substitute $k=4$and $L_2=3,L_3=4$in the given sequence.

$$\begin{gathered} L_4=L_{4-2}+L_{4-1} \\ =L_2+L_3 \\ =3+4 \\ =7 \end{gathered}$$

Step 3. Substitute k=5 and L3=4,L4=7 in the given sequence.

$$\begin{gathered} L_5=L_{5-2}+L_{5-1} \\ =L_3+L_4 \\ =4+7 \\ =11 \end{gathered}$$

Substitute $k=6$and $L_4=7,L_5=11$in the given sequence.

$$\begin{gathered} L_6=L_{6-2}+L_{6-1} \\ =L_4+L_5 \\ =7+11 \\ =18 \end{gathered}$$