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