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?

  • 1
    $\begingroup$ Have you tried a few simple cases to see the pattern? $\endgroup$
    – rtybase
    Sep 5, 2019 at 20:04
  • 1
    $\begingroup$ have you heard of induction? $\endgroup$
    – ggg
    Sep 5, 2019 at 20:06
  • $\begingroup$ Check my answer, I think it is easy; the seventh number in the list multiplied by 11. $\endgroup$ Sep 5, 2019 at 21:55

5 Answers 5



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

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


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.

  • 11
    $\begingroup$ There's no need for the teacher to calculate $5a+8b$ - it's already written on the board! $\endgroup$
    – Carmeister
    Sep 6, 2019 at 4:27
  • $\begingroup$ Actually the teacher can calculate $5a+8b$ essentially faster than it will be written on the board, and have additional time to perform multiplication by $11$ (and check the answer twice). $\endgroup$
    – Oleg567
    Sep 7, 2019 at 9:14

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:











The sum of those numbers is just $248$ (which is the $7^\text{th}$ number) $\times 11=2728$.

The rule is: $\boxed{7^\text{th}\text{ number }\times 11}$


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 $13*2 + 8 - 1 = 33$


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 $F_{10}x + (F_{11}-1)y$


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.

  • $\begingroup$ Thanks everyone! $\endgroup$ Sep 6, 2019 at 11:45

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