Difference between revisions 3684387 and 3684389 on mswiki

{{pelbagai isu|{{cleanup|reason=memerlukan penterjemahan segera kerana sudah ditinggalkan sejak tahun 2008|date=Ogos 2014}}{{Terjemah|en|fabonacci number|date=Ogos 2014}}}}
{{proses|BukanTeamBiasa}}
[[Image:FibonacciBlocks.svg|thumb|180px|right|Suatu ubinan dengan segi empat yang tepinya adalah nombor Fibonaci berturut-turut pada panjangnya]]
(contracted; show full)
:<math>p=29 \equiv 1 \pmod 4, \;\;(\tfrac{29}{5}) = +1, \frac{5(\frac{29}{5})-5}{2} =0\mbox{ and }\frac{5(\frac{29}{5})+5}{2}=5.</math>  
::<math>F_{14}=377 \mbox{ and } F_{15}=610.</math> 

::<math>5F_{14}^2=710645\equiv 0 \pmod {29} \;\;\mbox{ and }\;\;5F_{15}^2=1860500\equiv 5 \pmod {29}</math>
</blockquote>

===
Divisibility byKebolehbahagian dengan 11===
<math>\sum_{k=n}^{n+9} F_{k} = 11 F_{n+6}</math>

<blockquote>For example, letSebagai contoh, jadi ''n'' = 1:
<br>
F<sub>1</sub>+F<sub>2</sub>+...+F<sub>10</sub> = 1 + 1 + 2 + 3 + 5 + 8 + 13 + 21 + 34 + 55 = 143 = 11×13
<br>
''n'' = 2:
<br>
F<sub>2</sub>+F<sub>3</sub>+...+F<sub>11</sub> = 1 + 2 + 3 + 5 + 8 + 13 + 21 + 34 + 55 + 89 = 231 = 11×21
<br>
''n'' = 3:
<br>
F<sub>3</sub>+F<sub>4</sub>+...+F<sub>12</sub> = 2 +  3 + 5 + 8 + 13 + 21 + 34 + 55 + 89 + 144= 374 = 11×34

</blockquote>

In fact, the identity is true for allMalah, identitinya adalah benar bagi semua integers ''n'', not just positive onesbukan hanya yang positif:
<blockquote>
''n'' = 0:

<br>
F<sub>0</sub>+F<sub>1</sub>+...+F<sub>9</sub> = 0 + 1 + 1 + 2 + 3 + 5 + 8 + 13 + 21 + 34 = 88 = 11×8
<br>
''n'' = &minus;1:
(contracted; show full)
*[http://web.archive.org/web/20070715032716/http://mathdl.maa.org/convergence/1/?pa=content&sa=viewDocument&nodeId=630&bodyId=1002 Fibonacci Numbers] at [http://web.archive.org/web/20060212072618/http://mathdl.maa.org/convergence/1/ Convergence]
* [http://www.tools4noobs.com/online_tools/fibonacci/ Online Fibonacci calculator]

[[Kategori:Fibonacci numbers|*]]
[[Kategori:Articles containing proofs]]

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