Ytukyg

Why is 11 times the 7th term of a fibonacci series equal to the sum of 10 terms?

I was watching scam-school on youtube the other day and this number trick just astonished me. Can someone please explain why this works?

After a lot of searching, I've been stumbling onto slightly complicated mathematical explanations. An explanation of a simpler nature, one that a child can understand, would be much appreciated.

Also, Can you extend this to find the sum of n terms of a fibonacci type sequence?

@Claude Leibovici

In fact, there is a different way to answer this question using characteristic polynomials.

All Fibonacci-like sequences are associated with the same characteristic polynomial $x^2-x-1$ due to their common property : $$psi_{n+2}-psi_{n+1}-psi_{n}=0.$$

Let us define a new sequence in the following way :
$$chi_n:=(psi_{n+1}+psi_{n+2}+...+psi_{n+10})-11 psi_{n+7}. tag{1}$$

We want to show that, for any $n geq 0$, $chi_n=0$.

This is an easy consequence of the fact that the characteristic polynomial of sequence $chi_n$, i.e.,

$$p(x):=(x+x^2+...+x^9+x^{10})-11x^7$$

is divisible by $(x^2-x-1).$

Precisely :

$$p(x)=x(x^2 - x - 1)(x^7 + 2x^6 + 4x^5 - 4x^4 + x^3 - 2x^2 - 1).$$

As far as I know, it seems to be nothing more than coincidence. Say you have your starting numbers, $a$ and $b$. Your ten terms are
$a,b,a+b,a+2b,2a+3b,3a+5b,5a+8b,8a+13b,13a+21b,21a+34b$

the sum of which is $55a+88b$, which just happens to $11$ times the seventh term in your sequence.

If $F_n$ is the $n$th Fibonacci number (that is, $F_0=0$, $F_1=1$ and $F_{n+2}=F_{n+1}+F_n$), you can prove by induction that

$$sum_{k=0}^n F_k = F_{n+2}-1$$

It's obviously true for $n=0$, and if it is true for $n$, then

$$sum_{k=0}^{n+1} F_k = F_{n+1} + sum_{k=0}^{n} F_k = F_{n+1} +F_{n+2}-1 = F_{n+3}-1$$

Thus it's true for all $n$.

Your numerical trick is thus simply $143=F_{12} - 1 = 11 cdot F_7$. But notice that in general, $n not | F_{n+1} - 1$. For example, $609=F_{15}-1$, which is odd, thus not divisible by $14$.

You can also check for which values of $n$ it happens that $n|F_{n+1}-1$: $1, 4, 6, 9, 11, 19, 24, 29, 31, 34, 41, 46, 48, 59, 61, 71, 72, 79, 89, 94, 96, 100...$

This is sequence A219612 in OEIS, but if there is a pattern, it's not obvious.

As a follow-up, if you have a look at prime numbers in the preceding list, you get
$11, 19, 29, 31, 41, 59, 61, 71, 79, 89, 101, 109, 131...$

Apparently, they are primes congruent to $pm1$ modulo 5 (see OEIS A045468), but I don't have a proof.

If true, it would mean that for a prime $p$,

$$p | F_{p+1}-1 iff p equiv pm 1 pmod 5$$

It is unclear what you mean by a fibonacci series, but presumably is it simply one with $a_{n+1}=a_{n}+a_{n-1}$ and no particular starting points. Then you have in general

Sum of first ... terms   is ... times   ...th term

             (1)            (1)         (1) 

              2              1           3

             (3)            (2)         (3)   

              6              4           5

             10             11           7 

             14             29           9

            4n-2       int(phi^(2n+1))  2n+1 

where phi is $phi=frac{1+sqrt{5}}{2}$.

The most simple answer is not pretty but very simple:
Let use many times the fact that Un+2 = Un +Un+1

• U1+U2+U3+U4+U5+U6+U7+U8+U9+U10 =


• U1+U2+U3+U4+U5+U6+U7+2xU8+2xU9 = //Split of U10 in U9+U8


• U1+U2+U3+U4+U5+U6+3xU7+4xU8 = //Split of U9 in U8+U7


• U1+U2+U3+U4+U5+5xU6+7xU7 = //Split of U8 in U7+U6


• U1+U2+U3+U4+4xU6+8xU7 = //Merge of U5+U6 in U7


• U1+U2+U5+4xU6+8xU7 = //Merge of U3+U4 in U5


• U1+U2+3xU6+9xU7 = //Merge of U5+U6 in U7


• U3+3xU6+9xU7 = //Merge of U1+U2 in U3


• U3+U4+U5+2xU6+9xU7 = //Split of U6 in U5+U4


• U3+U4+U6+10xU7 = //Merge of U5+U6 in U7


• U5+U6+10xU7 = //Merge of U3+U4 in U5


• 11xU7 = //Merge of U5+U6 in U7

If the first term is x, and the second term is y, then
(1) x
(2) y
(3) x+ y
(4) x + 2y
(5) 2x + 3y
(6) 3x +5y
(7) 5x +8y
(8) 8x + 13y
(9) 13x + 21y
(10) 21x + 34y

The sum of all of those numbers is 55x + 88y, which factors to 11(5x + 8y)
Notice that the 7th term is 5x + 8y, so multiply the 7th term times 11 and you will get the sum of all the numbers in the sequence.