Alternative proof of the Fibonacci property $\sum\limits_{j=n}^{n+9}F_j=11\cdot F_{n+6}.$ If we consider the standard Fibonacci sequence$$F_0=0;\quad F_1=1;\quad F_n=F_{n-1}+F_{n-2}\quad\forall n\geq 2,$$ it is easy to proof that, given any chunk of $10$ consecutive Fibonacci, its sum is equal to $11$ times the seventh (this statement is also valid for any Fibonacci sequence startinng from any two values). That is $$\sum\limits_{j=n}^{n+9}F_j=11\cdot F_{n+6}$$ For example, lets consider the first $10$-numbers chunk of the sequence. Then $$0+1+1+2+3+5+\color{red}{8}+13+21+34=88=11\cdot\color{red}{8}.$$
I proved this by applying the recursive property several times in the following way:
\begin{gather*}\underbrace{F_n+F_{n+1}}_{F_{n+2}}+\underbrace{F_{n+2}+F_{n+3}}_{F_{n+4}}+\underbrace{F_{n+4}+F_{n+5}}_{F_{n+6}}+F_{n+6}+\underbrace{F_{n+7}}_{\underbrace{F_{n+5}}_{F_{n+3}+F_{n+4}}+F_{n+6}}+\underbrace{F_{n+8}}_{F_{n+6}+\underbrace{F_{n+7}}_{F_{n+5}+F_{n+6}}}+\underbrace{F_{n+9}}_{\underbrace{F_{n+7}}_{F_{n+5}+F_{n+6}}+\underbrace{F_{n+8}}_{F_{n+6}+F_{n+7}}}=&\\= F_{n+2}+F_{n+4}+F_{n+6}+F_{n+6}+F_{n+3}+F_{n+4}+F_{n+6}+F_{n+6}+F_{n+5}+F_{n+6}+F_{n+5}+F_{n+6}+F_{n+6}+F_{n+7}=&\\=\underbrace{F_{n+2}+F_{n+3}}_{F_{n+4}}+\underbrace{2F_{n+4}+2F_{n+5}}_{2F_{n+6}}+7F_{n+6}+\underbrace{F_{n+7}}_{F_{n+5}+F_{n+6}}=&\\=\underbrace{F_{n+4}+F_{n+5}}_{F_{n+6}}+10F_{n+6}=&\\=11F_{n+6}&\end{gather*}
This proof looks a bit naive to me compared to the strength of the statement. That is why I wanted to know if someone knows an alternative way to prove this statement. Thank you.

After OP changes 10 to 9.
Use $F_j=\frac{a^j-b^j}{\sqrt{5}}, a+b=1, ab=-1, a^2=a+1, b^2=b+1$
$$S_n=\sum_{j=n}^{n+9} F_j= \sum \frac{a^j-b^{j}}{\sqrt{5}}$$
USE sum of finite GP, changing the index as $j-n=k$
$$S_n=\sum_{k=0}^{9} \frac{a^{n+k}-b^{n+k}}{\sqrt{5}}=\frac{1}{\sqrt{5}} \left(a^n \frac{a^{10}-1}{a-1}-b^n\frac{b^{10}-1}{b-1}\right)$$
See $a-1=-b, -1/b=a$, then
$$S_n=\frac{1}{\sqrt{5}} [(a^{n+11}-a^{n+1})-(b^{n+11}-b^n)]=F_{n+11}-F_{n+1}.$$
I hope to come back.
Edit:
Now I prove that $F_{n+11}-F_{n+1}=11F_{n+6}$
Note that $$\left(\frac{1\pm\sqrt{5}}{2}\right)^5=\frac{11\pm5\sqrt{5}}{2}$$
$$F_{n+11}-F_{n+1}-11F_{n+6}=a(a^{10}-11a^5-1)-b(b^{10}-11b^5-1)~~~~(*)$$
As roots of $x^{10}-11x^5-1=0$ are $\frac{11\pm5\sqrt{5}}{2}=a^5,b^5;$
the RHS of (*) is zero.
