The sum of 8 consecutive Fibonacci numbers is not a Fibonacci number

Note that $F_{n+k}=F_{n+k+1}-F_{n+k-1}$ by summing this telescoping formula you get

$F_n+\cdots+F_{n+7}=F_{n+9}-F_{n+1}$

But $F_{n+1}<F_{n+7}$ is too small so this number is strictly greater than $F_{n+8}$ and also smaller than $F_{n+9}$.

We have that $$S:=F_n+\dots+F_{n+7}=F_{n+2}+F_{n+4}+F_{n+6}+F_{n+8}>F_{n+8}.$$ But on the other hand $$F_{n+9}=F_{n+8}+F_{n+7}=F_{n+8}+F_{n+6}+F_{n+5}=\dots\\=F_{n+8}+F_{n+6}+F_{n+4}+F_{n+2}+F_{n}+F_{n-1}=S+F_{n-1}>S.$$

In sum, this means that $F_{n+8}<S<F_{n+9}$.
But the Fibonacci sequence is streactly increasing, so there is no Fibonacci number between $F_{n+8}$ and $F_{n+9}$, i.e. $S$ is no Fibonacci number.

You can prove by induction that $$ F_n+\ldots+F_{n+7}=3\cdot L_{n+5}$$ and that $$ F_{n+8} < 3\cdot L_{n+5} < F_{n+9} $$ holds for any $n\geq 0$.
$L_n$ stands for the $n$-th Lucas number, $L_n=\varphi^n+\overline{\varphi}^n$.