Need help figure out a Fibonacci related math trick

Try it algebraically starting with $a$ and $b$ \begin{eqnarray*} a,b,a+b,a+2b,2a+3b,3a+5b,5a+8b, \\ 8a+13b,13a+21b,21a+34b. \end{eqnarray*} Now add these together and we get $55a+88b=11 (5a+8b)$.

So I guess your teacher took the first value multiplied by $5$ and added it to the second value multiplied by $8$ and then multiplied by $11$. Your teacher would have had plenty of time to do this calculation while then values were being added.

Hint:

$\begin{array}{rl} F(1) &= \color{blue}{F(3)}-F(2)\\ F(2)&= F(4)\color{blue}{-F(3)}\\ F(3)&=\color{red}{F(5)}-F(4)\\F(4)&=F(6)\color{red}{-F(5)}\\ \vdots\end{array}$

$F(1)+F(2)+\dots+F(n) = F(n+2)-F(2)$

That is because Fibonacci numbers have a number of properties, one of them being:

$$\sum_{i=0}^nF_i = F_{n+2} - 1 = 2F_n + F_{n-1} - 1$$

Proof is by induction

Hence, if the numbers are $0,1,1,2,3,5,8,13$, the sum will be $13*2 + 8 - 1 = 33$