I was wondering if anyone has any insights regarding the fact that the sum of any $a_1, \dots, a_{10}$ consecutive Fibonacci numbers is divisible by $11$ (and furthermore equals to $a_7*11$). What can this tell us about the series as a whole (its structure etc)?


Just write every term in the sum in terms of $a_1$ and $a_2$ (keeping in mind that $a_{n+2}=a_n+a_{n+1}$): $$a_1+a_2+(a_1+a_2)+(a_1 + 2a_2)+(2a_1+3a_2)+(3a_1+5a_2)+(5a_1+8a_2)+(8a_1+13a_2)+(13a_1+21a_2)+(21a_1+34a_2). $$

Then the sum is clearly equal to $55a_1+88a_2 = 11(5a_1+8a_2)$, which is $11$ times the seventh term of the sum.

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  • $\begingroup$ Yes, this shows us why its is divisible by 11 but not the meaning of it. what does it say about the structure etc. of the Fibonacci series? $\endgroup$ – NSChotzen Jul 23 '15 at 8:34
  • $\begingroup$ @NSChotzen I guess I answered the question as I read it in the title. I'm not sure this property can give us some powerful insight on this (otherwise very well studied) sequence. Perhaps questions of the form "What can X tell us about Y?" are slightly too vague, and in some cases the answer may very well be "Nothing much, really". $\endgroup$ – Pierre-Guy Plamondon Jul 23 '15 at 8:49
  • $\begingroup$ it seems to me that each set of a1,...,a10 acts like a telescopic expansion of the initial series: when (and I think only when) you accumulate 10 consecutive members, the whole group is the exact expansion of the 7th member. when a7=5a1+8a2 which in their turn of course, are comprised of 3a1+5a2 + 2a1+3a2 and in the new set it's 55a1+88a2 (which are comprised of 33a1+55a2 + 22a1 + 33a2). and it only happens in a set of 10. Right? $\endgroup$ – NSChotzen Jul 23 '15 at 9:59

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