{"id":98948,"date":"2018-03-01T18:05:49","date_gmt":"2018-03-01T21:05:49","guid":{"rendered":"http:\/\/clubes.obmep.org.br\/blog\/?page_id=98948"},"modified":"2024-04-04T04:28:22","modified_gmt":"2024-04-04T07:28:22","slug":"problema-para-ajudar-na-escola-uma-sequencia-numerica-disfarcada","status":"publish","type":"page","link":"https:\/\/clubes.obmep.org.br\/blog\/problema-para-ajudar-na-escola-uma-sequencia-numerica-disfarcada\/","title":{"rendered":".Problema para ajudar na escola: Uma sequ\u00eancia num\u00e9rica disfar\u00e7ada"},"content":{"rendered":"<p class=\"g2\" ><strong><a href=\"http:\/\/clubes.obmep.org.br\/blog\/?p=215560\" rel=\"noopener\" target=\"_blank\">&#9999; <span style=\"color:#A0522D;font-size:14px;\">Link do problema para dispositivos da <em>Apple<\/em>.<\/span><\/a><\/strong><\/p>\n<div class=\"g1n\" style=\"line-height:27px;\">\n<p class=\"tit\"  style=\"line-height:25px;margin:10px\">Problema <span style=\"font-size:12px; color:#0000FF\"><br \/>\n(A partir da 2\u00aa s\u00e9rie do E. M. &#8211; <span style=\"font-weight: normal;\"> N\u00edvel de dificuldade: M\u00e9dio<\/span>)<\/span><\/p>\n<hr style=\"height:1px; width:100%;background-color:#800000; margin-top: 10px; margin-bottom: 20px;border: 0 none\"\/>\nA sequ\u00eancia de n\u00fameros <\/p>\n<ul>\n<li>[tex]x_1 \\, , \\, x_2 \\, , \\, x_3, \\, \\cdots \\, [\/tex]<\/li>\n<\/ul>\n<p>est\u00e1 assim definida:<br \/>\n[tex]\\qquad \\qquad   \\begin{align*} x_1 &#038;=2\\\\<br \/>\nx_{n+1}&#038;=\\dfrac{x_n-1}{x_n+1},  \\text{ para cada inteiro positivo }n.\\end{align*}[\/tex]<\/p>\n<p>Determinar [tex]x_{2017}[\/tex].<\/p>\n<p style=\"margin:20px\">\n<\/div>\n<p style=\"margin:30px\">\n<div class=\"g1n\" style=\"line-height:30px;\">\n<p class=\"tit\" style=\"line-height:25px;margin:10px\">Solu\u00e7\u00e3o  <\/p>\n<hr style=\"height:1px; width:100%;background-color:#800000; margin-top: 10px; margin-bottom: 20px;border: 0 none\"\/>\nPara tentar entender a lei de forma\u00e7\u00e3o da sequ\u00eancia dada, vamos calcular alguns de seus termos iniciais:<\/p>\n<ul>\n<li>[tex]x_1 =2[\/tex]<\/li>\n<li>[tex] x_{2}=\\dfrac{x_1-1}{x_1+1}=\\dfrac{2-1}{2+1}=\\dfrac{1}{3}[\/tex]<\/li>\n<li>[tex] x_{3}=\\dfrac{x_2-1}{x_2+1}=\\dfrac{\\frac{1}{3}-1}{\\frac{1}{3}+1}=\\dfrac{\\frac{1-3}{3}}{\\frac{1+3}{3}}=\\dfrac{\\frac{-2}{\\cancel{3}}}{\\frac{4}{\\cancel{3}}}=-\\dfrac{2}{4}=-\\dfrac{1}{2}[\/tex]<\/li>\n<li>[tex] x_{4}=\\dfrac{x_3-1}{x_3+1}=\\dfrac{-\\frac{1}{2}-1}{-\\frac{1}{2}+1}=\\dfrac{\\frac{-1-2}{2}}{\\frac{-1+2}{2}}=\\dfrac{\\frac{-3}{\\cancel{2}}}{\\frac{1}{\\cancel{2}}}=-3[\/tex]<\/li>\n<li>[tex] x_{5}=\\dfrac{x_4-1}{x_4+1}=\\dfrac{-3-1}{-3+1}=\\dfrac{-4}{-2}=2.[\/tex]<\/li>\n<\/ul>\n<p>Como podemos observar que [tex]x_5=x_1[\/tex] e cada termo depende <strong>APENAS <\/strong> do seu termo imediatamente anterior, a nossa sequ\u00eancia \u00e9 peri\u00f3dica, com per\u00edodo [tex]4[\/tex]:<\/p>\n<ul>\n<li>[tex]\\textcolor{red}{x_1=2} \\, , \\, \\textcolor{blue}{x_2=\\dfrac{1}{3}} \\, , \\, \\textcolor{#32CD32}{x_3=-\\dfrac{1}{2}}, \\, \\textcolor{#FF00FF}{x_4=-3},[\/tex]\n<p style=\"margin:10px\">\n<p>[tex]\\textcolor{red}{x_5=2} \\, , \\, \\textcolor{blue}{x_6=\\dfrac{1}{3}} \\, , \\, \\textcolor{#32CD32}{x_7=-\\dfrac{1}{2}}, \\, \\textcolor{#FF00FF}{x_8=-3},[\/tex]<\/p>\n<p style=\"margin:10px\">\n<p>[tex]\\textcolor{red}{x_9=2} \\, , \\, \\textcolor{blue}{x_{10}=\\dfrac{1}{3}} \\, , \\, \\textcolor{#32CD32}{x_{11}=-\\dfrac{1}{2}}, \\, \\textcolor{#FF00FF}{x_{12}=-3}, \\, \\cdots[\/tex]<\/li>\n<\/ul>\n<p>Desta maneira, para determinarmos o valor do termo [tex]x_{2017} \\, [\/tex], basta dividirmos  [tex]2017[\/tex] por [tex]4[\/tex] e observarmos o resto da divis\u00e3o:<br \/>\n[tex]\\qquad \\qquad \\begin{array}{r}<br \/>\n2017 \\, \\end{array}   \\begin{array}{|r}<br \/>\n \\, 4 \\,  \\,  \\,  \\,  \\\\  \\hline<br \/>\n\\end{array}[\/tex]<br \/>\n[tex]\\qquad \\qquad\\begin{array}{r}<br \/>\n \\,  \\,  \\,  \\, \\boxed{1}<br \/>\n\\end{array}\\begin{array}{r}<br \/>\n  \\,  \\, 504<br \/>\n\\end{array}[\/tex]<\/p>\n<p style=\"margin:20px\">\n<p>Como [tex]2017[\/tex] \u00e9 um m\u00faltiplo de [tex]4[\/tex] mais [tex]1[\/tex], a nossa sequ\u00eancia segue:<\/p>\n<ul>\n<li>[tex]\\textcolor{red}{x_1=2} \\, , \\, \\textcolor{blue}{x_2=\\dfrac{1}{3}} \\, , \\, \\textcolor{#32CD32}{x_3=-\\dfrac{1}{2}}, \\, \\textcolor{#FF00FF}{x_4=-3},[\/tex]\n<p style=\"margin:10px\">\n<p>[tex]\\textcolor{red}{x_5=2} \\, , \\, \\textcolor{blue}{x_6=\\dfrac{1}{3}} \\, , \\, \\textcolor{#32CD32}{x_7=-\\dfrac{1}{2}}, \\, \\textcolor{#FF00FF}{x_8=-3},[\/tex]<\/p>\n<p style=\"margin:10px\">\n<p>[tex]\\textcolor{red}{x_9=2} \\, , \\, \\textcolor{blue}{x_{10}=\\dfrac{1}{3}} \\, , \\, \\textcolor{#32CD32}{x_{11}=-\\dfrac{1}{2}}, \\, \\textcolor{#FF00FF}{x_{12}=-3},[\/tex]<br \/>\n[tex]\\hspace{5cm} \\vdots[\/tex]<br \/>\n[tex]\\textcolor{red}{\\boxed{x_{2017}=2}} \\, , \\, \\cdots[\/tex] .\n<\/li>\n<\/ul>\n<hr style=\"height:1px; width:100%;background-color:#800000; margin-top: 10px; margin-bottom: 15px;border: 0 none\"\/>\n<p class=\"g2\" style=\"line-height:20px;\">Solu\u00e7\u00e3o elaborada pelos <strong>Moderadores do Blog<\/strong>.<\/p>\n<\/div>\n<p style=\"margin:20px\">\n<table style=\" height:auto; margin-left: auto; margin-right: auto;width:640px;\">\n<tr>\n<td class=\"bf\"><img loading=\"lazy\" decoding=\"async\" class=\"wp-image-1001 alignright\" style=\"border: 0px;\" alt=\"\" src=\"http:\/\/clubes.obmep.org.br\/blog\/wp-content\/uploads\/2013\/01\/menino.jpg\" width=\"80\" height=\"115\" \/><\/td>\n<td class=\"bf\">\n<p class=\"fala\" style=\"width:520px;background-color: #FFF5EE; padding:10px; font-family: calibri; font-size: 13px;color: #1C1C1C; line-height: 20px; \"> Se for conveniente, voc\u00ea pode obter um arquivo PDF desta p\u00e1gina, com o problema e a solu\u00e7\u00e3o, clicando no bot\u00e3o abaixo.<br \/>\nVoc\u00ea pode abrir o arquivo diretamente no seu navegador (<em>Chrome<\/em>, <em>Edge<\/em>, <em>Firefox<\/em>, <em>Safari<\/em>, entre outros), mas tamb\u00e9m pode utilizar o software gratuito <em><strong>Adobe Acrobat Reader<\/strong><\/em>.<br \/>\nCaso o dispositivo que voc\u00ea est\u00e1 utilizando n\u00e3o tenha o <strong><em>Acrobat Reader<\/em><\/strong> instalado, \u00e9 s\u00f3 clicar <a href=\"http:\/\/get.adobe.com\/br\/reader\/\" rel=\"noopener\" target=\"_blank\"><strong>AQUI<\/strong><\/a> para fazer o download adequado ao seu dispositivo.<\/p>\n<\/td>\n<\/tr>\n<\/table>\n<div style=\" float:right;  width: 100px;\">\n<p style=\"text-align: center; border-style: dotted; border-color :#FFFAFA; border-width :thin; width:90px;  height:40px;  padding: 5px; background-color: #800000;\">\n<a href=\"http:\/\/clubes.obmep.org.br\/blog\/wp-content\/uploads\/2019\/11\/Uma-sequ\u00eancia-num\u00e9rica-disfar\u00e7ada.pdf\" title=\" Arquivo PDF\" download=\"Clubes_ Uma sequ\u00eancia num\u00e9rica disfar\u00e7ada\"><span style=\"color: #FFFAFA; font-family: georgia; font-size: 16px;\"><strong>Arquivo <\/strong><\/span><\/a>\n <\/p>\n<\/div>\n","protected":false},"excerpt":{"rendered":"<p>&#9999; Link do problema para dispositivos da Apple. Problema (A partir da 2\u00aa s\u00e9rie do E. M. &#8211; N\u00edvel de dificuldade: M\u00e9dio) A sequ\u00eancia de n\u00fameros [tex]x_1 \\, , \\, x_2 \\, , \\, x_3, \\, \\cdots \\, [\/tex] est\u00e1 assim definida: [tex]\\qquad \\qquad \\begin{align*} x_1 &#038;=2\\\\ x_{n+1}&#038;=\\dfrac{x_n-1}{x_n+1}, \\text{ para cada inteiro positivo }n.\\end{align*}[\/tex] Determinar &hellip; <\/p>\n<p><a class=\"more-link btn\" href=\"https:\/\/clubes.obmep.org.br\/blog\/problema-para-ajudar-na-escola-uma-sequencia-numerica-disfarcada\/\">Continue lendo<\/a><\/p>\n","protected":false},"author":2,"featured_media":0,"parent":0,"menu_order":0,"comment_status":"closed","ping_status":"closed","template":"template-onecolumn.php","meta":{"footnotes":""},"class_list":["post-98948","page","type-page","status-publish","hentry","nodate","item-wrap"],"_links":{"self":[{"href":"https:\/\/clubes.obmep.org.br\/blog\/wp-json\/wp\/v2\/pages\/98948","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/clubes.obmep.org.br\/blog\/wp-json\/wp\/v2\/pages"}],"about":[{"href":"https:\/\/clubes.obmep.org.br\/blog\/wp-json\/wp\/v2\/types\/page"}],"author":[{"embeddable":true,"href":"https:\/\/clubes.obmep.org.br\/blog\/wp-json\/wp\/v2\/users\/2"}],"replies":[{"embeddable":true,"href":"https:\/\/clubes.obmep.org.br\/blog\/wp-json\/wp\/v2\/comments?post=98948"}],"version-history":[{"count":15,"href":"https:\/\/clubes.obmep.org.br\/blog\/wp-json\/wp\/v2\/pages\/98948\/revisions"}],"predecessor-version":[{"id":215564,"href":"https:\/\/clubes.obmep.org.br\/blog\/wp-json\/wp\/v2\/pages\/98948\/revisions\/215564"}],"wp:attachment":[{"href":"https:\/\/clubes.obmep.org.br\/blog\/wp-json\/wp\/v2\/media?parent=98948"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}