Difference between revisions 3684404 and 3684408 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>\sum_{n = 1}^{\infty}{\frac {F(n)}{10^{(k + 1)(n + 1)}}} = \frac {1}{10^{2k + 2} - 10^{k + 1} - 1}</math> untuk semua integer <math>k >= 0</math>. Secara bicara, :<math>\sum_{n=0}^\infty\,\frac{F_n}{k^{n}}\,=\,\frac{k}{k^{2}-k-1}.</math> == Reciprocal sumsJumlah salingan== <!-- {{cite book | last =Borwein | first =Jonathan M. | authorlink =Jonathan Borwein | coauthors =[[Peter Borwein|Peter B. Borwein]] | title =Pi and the AGM: A Study in Analytic Number Theory and Computational Complexity | pages =91–101 | publisher =Wiley | year =1998 | month =July | url =http://www.wiley.com/WileyCDA/WileyTitle/productCd-047131515X.html | id = ISBN 978-0-471-31515-5 }} It credits some formulae to {{cite journal | author = Landau, E. | title = Sur la Série des Invers de Nombres de Fibonacci | journal = Bull. Soc. Math. France | volume = 27 | year = 1899 | pages = 298–300}} --> Infinite sums over reciprocal Fibonacci numbers can sometimes be evaluated in terms of [[theta functionJumlah tidak terhingga ke atas nombor Fibonacci salingan kadang-kadang boleh dinilai dari segi [[fungsi theta]]s. For example, we can write the sum of every odd-indexed reciprocal Fibonacci number asSebagai contoh, kita boleh menulis jumlah setiap nombor Fibonacci salingan indeks ganjil sebagai :<math>\sum_{k=0}^\infty \frac{1}{F_{2k+1}} = \frac{\sqrt{5}}{4}\vartheta_2^2 \left(0, \frac{3-\sqrt 5}{2}\right) ,</math> dand the sum of squared reciprocal jumlah kuasa dua nombor Fibonacci numbers assalingan :<math>\sum_{k=1}^\infty \frac{1}{F_k^2} = \frac{5}{24} \left(\vartheta_2^4\left(0, \frac{3-\sqrt 5}{2}\right) - \vartheta_4^4\left(0, \frac{3-\sqrt 5}{2}\right) + 1 \right).</math> If we add 1 to each Fibonacci number in the first sumJika kita menambah 1 untuk setiap nombor Fibonacci dalam jumlah yang pertama, there is also the closed formdapat juga bentuk tertutup :<math>\sum_{k=0}^\infty \frac{1}{1+F_{2k+1}} = \frac{\sqrt{5}}{2},</math> dand there is a nice ''nested'' sum of squared Fibonacci numbers giving the reciprocal of the [[golden ratio]], :<math>\sum_{k=1}^\infty \frac{(-1)^{k+1}}{\sum_{j=1}^k {F_{j}}^2} = \frac{\sqrt{5}-1}{2}.</math> Results such as these make it plausible that a closed formula for the plain sum of reciprocal Fibonacci numbers could be found, but none is yet known. Despite that, the [[reciprocal ada jumlah ''tersarang'' bagi kuasa dua nombor Fibonacci lalu memberi salingan [[nisbah emas]], :<math>\sum_{k=1}^\infty \frac{(-1)^{k+1}}{\sum_{j=1}^k {F_{j}}^2} = \frac{\sqrt{5}-1}{2}.</math> Keputusan seperti ini menjadikannya munasabah bahawa rumus tertutup untuk jumlah mendatar bagi nombor Fibonacci salingan boleh didapati, tetapi tiada yang masih diketahui. Walaupun begitu, [[pemalar Fibonacci constsalingant]] :<math>\psi = \sum_{k=1}^{\infty} \frac{1}{F_k} = 3.359885666243 \dots</math> has been proved [[irrational number|irrational]] bytelah dibuktikan [[nombor tak nisbah|tidak bernisbah]] oleh [[Richard André-Jeannin]]. ==Nombor perdana dan kebolehbahagian== {{main|Fibonacci perdana}} '''Fibonacci perdana''' adalah nombor Fibonacci yang [[nombor perdana|perdana ]] {{OEIS|id=A005478}}. The first few are: : 2, 3, 5, 13, 89, 233, 1597, 28657, 514229, … (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]] <!-- interwiki --> All content in the above text box is licensed under the Creative Commons Attribution-ShareAlike license Version 4 and was originally sourced from https://ms.wikipedia.org/w/index.php?diff=prev&oldid=3684408.
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