Q. 10 - Calculus

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 $L_6$?

Step-by-Step Solution

Answer

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

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 L 1 = 1 , L 2 = 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 L 3 = 4 , L 4 = 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.

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