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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?

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  • $\begingroup$ Is this what you are looking for ? Sum[Fibonacci[n], {n, 1, m}] = Fibonacci[2 + m] - 1 $\endgroup$ – Claude Leibovici Dec 9 '13 at 9:05
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@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 :

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

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  • $\begingroup$ This can be used to find further such coincidental properties, maybe even a general condition for some $n^{th}$ and $m^{th}$ Fibonacci terms to related. Thank you, Jean Marie. $\endgroup$ – Nick Dec 2 '18 at 8:28
  • $\begingroup$ @Nick You are right : the simple recipe is to take any multiple of polynomial $x^2-x-1$ ; but only those having some "appeal" are eligible for a problem like this one... $\endgroup$ – Jean Marie Dec 2 '18 at 13:29
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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.

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  • $\begingroup$ Can you extend that to find the sum of n fibonacci numbers? $\endgroup$ – Nick Dec 9 '13 at 8:15
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    $\begingroup$ Consider the regular Fibonacci sequence: 1,1,2,3,5 etc. To find the sum of the first n terms, take the n+2nd term, and subtract 1. For example, to sum 1+1+2+3+5+8+13+21, go 2 terms over, to 55, and subtract 1, to get 54. In general, if you have a Fibonacci-like sequence starting with a and b instead of 1 and 1, to find the desired sum, move two over, and subtract b instead of 1. $\endgroup$ – ant11 Dec 9 '13 at 8:20
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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 $p$,

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

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

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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
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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.

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    $\begingroup$ Hello and welcome to math.stackexchange! It is great that you want to start answering questions, but this particular question was asked a year ago and already has an answer that is virtually identical to yours. You can find more recent and unanswerd questions on the front page and by clicking on “unanswered” below the logo. (You might also want to look at the editing guide to learn how you can use MathJax to make your formulas look prettier.) $\endgroup$ – Eike Schulte Nov 24 '15 at 2:11

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