# Show that f_n−1+L_n=2f_n.So we need to find a 2 to 1 correspondence. Set 1: Tilings an n-board. Set 2: Tiling of an n−1-board or tiling of an 𝑛-bracelet. So we need to decompose a tiling of an 𝑛-board to a tiling of an n−1-board or a tiling of an n−1-bracelet?

Show that ${f}_{n-1}+{L}_{n}=2{f}_{n}$.
So we need to find a $2$ to $1$ correspondence.
Set 1: Tilings an $n$-board.
Set 2: Tiling of an $n-1$-board or tiling of an 𝑛-bracelet.
So we need to decompose a tiling of an 𝑛-board to a tiling of an $n-1$-board or a tiling of an $n-1$-bracelet?
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Marshall Horne
${f}_{n}$ denotes number of tilings of the $n×1$ board by $2×1$ and $1×1$ pieces.
${l}_{n}$ is almost the same thing, but the ends of the board is joined so that they form a kind of "bracelet" (i.e., one domino can cover the first and the last position; since they are adjacent now.) Also it should be mentioned that the position where the ends are glued together is fixed.
The above gives a combinatorial interpretation of Fibonacci and Lucas numbers, since ${f}_{n}={F}_{n+1}$ and ${l}_{n}={L}_{n}$.
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Domianpv
Your identity is equivalent to ${L}_{n}=2{f}_{n}-{f}_{n-1}={f}_{n}+{f}_{n-2}$ (using ${f}_{n}-{f}_{n-1}={f}_{n-2}$). A combinatorial proof of this identity is easy (and it is given in the book you quote as identity $32$).
Basically the same proof, just formulated another way, is rewriting the identity as ${f}_{n}-{f}_{n-1}={L}_{n}-{f}_{n}$ and to check that both expressions are equal to the number of tilings of $\left(n-2\right)$-board.