Cassini's identity and sums of Fibonacci squares

1. By induction, $M^n=\begin{pmatrix}F_{n+1}&F_n\\F_n&F_{n-1}\end{pmatrix}$. Taking determinants, $\det M^n=(\det M)^n=(-1)^n$, i.e. $F_{n+1}F_{n-1}-F_n^2=(-1)^n$.

2. For $n=1$: $F_1^2=1=F_1F_2$. If the formula holds for $n$, then
$$F_nF_{n+1}+F_{n+1}^2=F_{n+1}(F_n+F_{n+1})=F_{n+1}F_{n+2}.$$

3. $M^{2n+1}=M^nM^{n+1}$, and the $(1,2)$ entry of $M^k$ is $F_k$. The $(1,2)$ entry of the product is $F_{n+1}\cdot F_{n+1}+F_n\cdot F_n$. Hence $F_{2n+1}=F_n^2+F_{n+1}^2$.

Computation. $F_{10}=55$, $F_{11}=89$, so $\sum_{k=1}^{10}F_k^2=55\cdot89=4895$.

Lesson. Powers of the companion matrix $M$ package the Fibonacci recurrence: determinants and products of powers give identities for free.