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| largest term in sequence https://forumdematematica.org/viewtopic.php?f=21&t=1974 |
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| Autor: | kinu [ 09 mar 2013, 07:11 ] |
| Título da Pergunta: | largest term in sequence |
The Largest term in the Sequence \(\displaystyle \frac{1}{503}\;,\frac{4}{524}\;,\frac{9}{581}\;,\frac{16}{692},.....\) is |
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| Autor: | Fraol [ 09 mar 2013, 14:06 ] |
| Título da Pergunta: | Re: largest term in sequence |
Good morning, The main difficulty in this problem was to find the general term of the sequence. Especially in relation to the denominator. After some effort we have the following: \(a_n = \frac{n^2}{500 + 3n^3}\) Clearly, when \(n\) tends to infinity then \(a_n\) tends to \(0\) so \((a_n)\) have a largest term. So we must derivate \(a_n\): \(a'_n = \frac{2n(500+3n^3)-n^2(9n^2)}{(500+3n^3)2}\) Now we set this result to zero to find \(n\): \(a'_n = \frac{2n(500+3n^3)-n^2(9n^2)}{(500+3n^3)2} = 0 <=> 2n(500+3n^3)-n^2(9n^2) = 0 <=> 2n(500+3n^3) = n^2(9n^2)\) So \(1000 + 6n^3 = 9n^3 <=> n^3 = \frac{1000}{3} <=> n \approx 6,9\), which leads to \(n = 7\) and \(a_7 = \frac{7^2}{(500+3\cdot7^3)}\) is the largest term in the sequence. |
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| Autor: | Fraol [ 09 mar 2013, 14:15 ] |
| Título da Pergunta: | Re: largest term in sequence |
Hi, To verify the above results, considering that the general term formula is correct, I tabulated the first 15 terms of the sequence: Código: n a_n 1 0,0019881 2 0,0076336 3 0,0154905 4 0,0231214 5 0,0285714 6 0,0313589 7 0,0320471 8 0,0314342 9 0,0301451 10 0,0285714 11 0,0269308 12 0,0253343 13 0,0238330 14 0,0224462 15 0,0211765 . |
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