Need help figure out a Fibonacci related math trick My math teacher used to do a trick where he would have a student write $2$ numbers on the board then add the first to the second to create the third then add the second to the third and so on until there were $10$ numbers. He would then turn around and add them up in $2$ seconds. How did he do this?
 A: Hint:
$\begin{array}{rl} F(1) &= \color{blue}{F(3)}-F(2)\\
F(2)&= F(4)\color{blue}{-F(3)}\\
F(3)&=\color{red}{F(5)}-F(4)\\F(4)&=F(6)\color{red}{-F(5)}\\
\vdots\end{array}$

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

A: 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.
A: 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 $13*2 + 8 - 1 = 33$
A: Multiplying any natural number by $11$ is so easy, check here.
Now the solution for your problem is to multiply the $7^\text{th}$ number in the list by $11$

Have this example: our first two numbers are $16$ and $21$
So the list is:
$16$
$21$
$37$
$58$
$95$
$153$
$248$
$401$
$649$
$1050$
The sum of those numbers is just $248$ (which is the $7^\text{th}$ number) $\times 11=2728$.

The rule is: $\boxed{7^\text{th}\text{ number }\times 11}$
A: Well, to answer the question as to how he did it:  If the first number is $x$ and the second number is $y$ then every other number and the sum of all ten numbers will a combination of $x$ and $y$.  As you do the same thing every time the final sum will be the same combination.  Your teacher merely memorized that the final sum would be $55x + 88y$.
As to how we would know the final number is $55x+88y$ we can
1) Simply do it.  The ten numbers are $x,y,x+y, x+2y, 2x+3y, 3x+5y,5x+8y,8x+13y,13x+21y, 21x+34y$ and the sum is $55x+88y$.
2) Try to find a way  to generalize this without doing each sum.
We notice the number of $x$s involved are $1,0,1,1, 2,etc.$.  After a slow start once we have $1,1$ this has to follow the Fibonacci sequence.  So if the $k$th number is $a_k x + b_k y$ we know $a_k= F_{k-2}$, the $k$th fibonacci number.
We notice the number of $y$s involved are $0,1,1,2, etc.$ and those are the Fibonacci numbers too but with a quicker start.  $b_k = F_{k-1}$.
So the $k$th number is $F_{k-2}x + F_{k-1}y$.
The final total after ten numbers is therefore $(1 +\sum_{k=1}^8 F_k)x + (\sum_{k=1}^9 F_k) y$.
There's an interesting formula that 
$\sum_{k=1}^n F_k =  F_{n+2} - 1$.
So the sum is $F_{10}x + (F_{11}-1)y$
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Another answer states that the answer is that the sum is the $7$th number times $11$.
So does $11(F_{5}x + F_{6}y) = F_{10}x + (F_{11} -1)y$?
Well, $11(F_5x + F_6y) = 11(5x + 8y)=55x + 88y = 55x + (89-1)y$.
The Fibonacci series is $1,1,2,3,5,8,13,21,34,55,89$ so,  yes, indeed this is true.
So that's actually how the teacher did it so quickly.  You wrote down the $7$th term and he multiplied it by $11$ in his head.
