Relating Fibonacci numbers to binomial coefficient

Question

Let $F_n$ denote the $n^{\text{th}}$ Fibonacci number, show that $$F_{2n+2}=\sum_{i+j \leq n}\binom{n-i}{j}\binom{n-j}{i}$$

Answer 1

We can rewrite the summation and apply this identity:

$$\begin{align*} \sum_{i+j\le n}\binom{n-i}j\binom{n-j}i&=\sum_{k=0}^n\sum_{i=0}^k\binom{n-i}{k-i}\binom{n-k+i}i\\ &=\sum_{k=0}^n\sum_{i=0}^k\binom{n-i}{n-k}\binom{n-k+i}{n-k}\\ &=\sum_{k=0}^n\binom{2n-k+1}{2n-2k+1}\\ &=\sum_{k=0}^n\binom{2n-k+1}k \end{align*}$$

Now use the following familiar Fibonacci identity, which is easily proved by induction on $n$:

$$F_n=\sum_{k=0}^{\lfloor\frac{n-1}2\rfloor}\binom{n-k-1}k$$

Answer 2

A combinatorial approach for the last part of the accepted answer.

Let $A_n$ be the number of strings over the alphabet $\Sigma=\{0,1\}$ in which there are no adjacent $1$s. A string fulfilling these constraints can only start with $0$ or $10$, such that $A_{n+2}=A_{n+1}+A_n$. Since $A_1=2$ and $A_2=3$, we have $A_n=F_{n+2}$, so $F_{2n+2}$ is the number of constrained strings with $2n$ characters, containing at most $n$ characters $1$. If we assume there are exactly $k$ characters $1$, by removing them we get a string of zeroes with length $2n-k$. If we insert back $k$ characters $1$ we have $2n+1-k$ allowed positions, between adjacent characters zero, or at the beginning/end of the string. It follows that

$$ F_{2n+2}=\sum_{k=0}^{n}\binom{2n+1-k}{k}. $$