{"id":34579,"date":"2015-04-08T16:27:07","date_gmt":"2015-04-08T19:27:07","guid":{"rendered":"http:\/\/clubes.obmep.org.br\/blog\/?page_id=34579"},"modified":"2023-03-21T22:06:18","modified_gmt":"2023-03-22T01:06:18","slug":"problema-de-gincana-area-2","status":"publish","type":"page","link":"https:\/\/clubes.obmep.org.br\/blog\/problema-de-gincana-area-2\/","title":{"rendered":".Problema de Gincana: Uma soma de fra\u00e7\u00f5es"},"content":{"rendered":"<p style=\"margin:10px\">\n<div class=\"g1n\">\n<p class=\"tit\" style=\"line-height:25px;margin:10px\">Problema<\/p>\n<hr style=\"height:1px; width:100%;background-color:#800000; margin-top: 10px; margin-bottom: 10px;border: 0 none\"\/>\n<p style=\"margin:20px\">\n<p>Sabendo que<br \/>\n[tex]\\qquad \\dfrac{1}{10}-\\dfrac{1}{11}=\\dfrac{11-10}{10 \\times 11}=\\dfrac{1}{10 \\times 11}[\/tex],<br \/>\no valor da soma<br \/>\n[tex]\\quad S=\\dfrac{1}{10 \\times 11}+\\dfrac{1}{11 \\times 12}+ \\ldots +\\dfrac{1}{19 \\times 20}[\/tex]<br \/>\n\u00e9:<\/p>\n<p>A) [tex]\\dfrac{1}{10}[\/tex].<\/p>\n<p>B) [tex]\\dfrac{5}{20}[\/tex].<\/p>\n<p>C) [tex]\\dfrac{2}{5}[\/tex].<\/p>\n<p>D) [tex]\\dfrac{1}{20}[\/tex].<\/p>\n<p>E) Nenhuma das respostas anteriores.<\/p>\n<p style=\"margin:20px\">\n<\/div>\n<p style=\"margin:40px\">\n<div class=\"g1n\">\n<p class=\"tit\" style=\"line-height:30px;margin:10px\">Solu\u00e7\u00e3o<\/p>\n<hr style=\"height:1px; width:100%;background-color:#800000; margin-top: 10px; margin-bottom: 10px;border: 0 none\"\/>\n<p style=\"margin:20px\">\n<p>Podemos generalizar o racioc\u00ednio explicitado na hip\u00f3tese. Assim, se [tex]n[\/tex] \u00e9 um inteiro positivo, temos que:<\/p>\n<p style=\"line-height: 55px;\">\n[tex]\\quad\\qquad \\dfrac{1}{n}-\\dfrac{1}{n+1}=\\dfrac{(n+1)-n}{n \\times (n+1)}=\\dfrac{1}{n \\times (n+1)}[\/tex].<br \/>\nDessa forma, aplicando a igualdade obtida em cada parcela da soma [tex]S[\/tex], segue que:<br \/>\n[tex]\\quad S=\\dfrac{1}{10 \\times 11}+\\dfrac{1}{11 \\times 12}+ \\ldots +\\dfrac{1}{19 \\times 20}[\/tex]<br \/>\n[tex]\\quad S=\\left[\\dfrac{1}{10}-\\dfrac{1}{11}\\right]+\\left[\\dfrac{1}{11}-\\dfrac{1}{12}\\right]+\\left[\\dfrac{1}{12}-\\dfrac{1}{13}\\right]+ \\ldots +\\left[\\dfrac{1}{19}-\\dfrac{1}{20}\\right][\/tex]<br \/>\n[tex]\\quad S=\\dfrac{1}{10}+\\left[-\\dfrac{1}{11}+\\dfrac{1}{11}\\right]+\\left[-\\dfrac{1}{12}+\\dfrac{1}{12}\\right]+ \\ldots +\\left[-\\dfrac{1}{19}+\\dfrac{1}{19}\\right]-\\dfrac{1}{20}[\/tex]<br \/>\n[tex]\\quad S=\\dfrac{1}{10}+0+0+ \\ldots +0-\\dfrac{1}{20}[\/tex]<br \/>\n[tex]\\quad S=\\dfrac{1}{10}-\\dfrac{1}{20}[\/tex]<br \/>\n[tex]\\quad S=\\dfrac{20-10}{10\\times 20}[\/tex]<br \/>\n[tex]\\quad S=\\dfrac{10}{200}[\/tex]<br \/>\n[tex]\\quad S=\\dfrac{1}{20}[\/tex].<\/p>\n<p>Portanto:<br \/>\n [tex]\\qquad\\qquad S=\\dfrac{1}{10 \\times 11}+\\dfrac{1}{11 \\times 12}+ \\ldots +\\dfrac{1}{19 \\times 20}=\\dfrac{1}{20}.[\/tex]<\/p>\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:30px\">\n<div class=\"g3\" style=\"width: 480px;\">\nSegunda Gincana de 2014 &#8211; Clubes de Matem\u00e1tica da OBMEP<br \/>\nN\u00edvel A &#8211; Quest\u00e3o Dif\u00edcil<\/div>\n","protected":false},"excerpt":{"rendered":"<p>Problema Sabendo que [tex]\\qquad \\dfrac{1}{10}-\\dfrac{1}{11}=\\dfrac{11-10}{10 \\times 11}=\\dfrac{1}{10 \\times 11}[\/tex], o valor da soma [tex]\\quad S=\\dfrac{1}{10 \\times 11}+\\dfrac{1}{11 \\times 12}+ \\ldots +\\dfrac{1}{19 \\times 20}[\/tex] \u00e9: A) [tex]\\dfrac{1}{10}[\/tex]. B) [tex]\\dfrac{5}{20}[\/tex]. C) [tex]\\dfrac{2}{5}[\/tex]. D) [tex]\\dfrac{1}{20}[\/tex]. E) Nenhuma das respostas anteriores. Solu\u00e7\u00e3o Podemos generalizar o racioc\u00ednio explicitado na hip\u00f3tese. Assim, se [tex]n[\/tex] \u00e9 um inteiro positivo, temos que: [tex]\\quad\\qquad &hellip; <\/p>\n<p><a class=\"more-link btn\" href=\"https:\/\/clubes.obmep.org.br\/blog\/problema-de-gincana-area-2\/\">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-34579","page","type-page","status-publish","hentry","nodate","item-wrap"],"_links":{"self":[{"href":"https:\/\/clubes.obmep.org.br\/blog\/wp-json\/wp\/v2\/pages\/34579","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=34579"}],"version-history":[{"count":65,"href":"https:\/\/clubes.obmep.org.br\/blog\/wp-json\/wp\/v2\/pages\/34579\/revisions"}],"predecessor-version":[{"id":185097,"href":"https:\/\/clubes.obmep.org.br\/blog\/wp-json\/wp\/v2\/pages\/34579\/revisions\/185097"}],"wp:attachment":[{"href":"https:\/\/clubes.obmep.org.br\/blog\/wp-json\/wp\/v2\/media?parent=34579"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}