$\bullet~~$ If $x^2=x+1$ ,$n\ge 2$ then we have $x^n=F_nx+F_{n-1}$, where $F_n$ is the $n^{th}$ Fibonacci number.

$\bullet~~\sum_{i=2}^{n} \tau^i+(1-\tau)^i =3(F_{n+1}-1)$ where $F_n$ is the $n$th Fibonacci number and $\tau$ is the Golden ratio. It follows from the identity stated in the post by David.

Here you can get a full proof for both.

$\bullet~~$ Recently I found a generalized version for Cesaro theorem [theorem no. 81] in Fibonacci numbers. It states that - for fixed $p$ show that,$$\sum_{k=1}^{n}\dbinom{n}{k}F_p^kF_{p-1}^{n-k}F_k=F_{pn}.$$ You can get a full proof in Issue 3, Mathematical Reflections, 2018 (page 16).

$\bullet~~$ If $x^2=x+1$ ,$n\ge 2$ then we have $x^n=F_nx+F_{n-1}$, where $F_n$ is the $n^{th}$ Fibonacci number.

$\bullet~~\sum_{i=2}^{n} \tau^i+(1-\tau)^i =3(F_{n+1}-1)$ where $F_n$ is the $n$th Fibonacci number and $\tau$ is the Golden ratio. It follows from the identity stated in the post by David.

Here you can get a full proof for both.

$\bullet~~$ Recently I found a generalized version for Cesaro theorem [theorem no. 81] in Fibonacci numbers. It states that - for fixed $p$,$$\sum_{k=1}^{n}\dbinom{n}{k}F_p^kF_{p-1}^{n-k}F_k=F_{pn}.$$ You can get a full proof in Issue 3, Mathematical Reflections, 2018 (page 16).