{"id":107024,"date":"2018-07-13T21:18:43","date_gmt":"2018-07-14T00:18:43","guid":{"rendered":"http:\/\/clubes.obmep.org.br\/blog\/?page_id=107024"},"modified":"2023-05-03T02:54:41","modified_gmt":"2023-05-03T05:54:41","slug":"problema-fracoes-produtos-e-somas","status":"publish","type":"page","link":"https:\/\/clubes.obmep.org.br\/blog\/problema-fracoes-produtos-e-somas\/","title":{"rendered":".Problema: Fra\u00e7\u00f5es, Produtos e Somas"},"content":{"rendered":"

\n

\n

Problema
\n (Indicado a partir do 9\u00ba ano do E. F.)<\/span><\/p>\n

\n

Calcule o valor da express\u00e3o:<\/p>\n

[tex]\\dfrac{1}{2\\cdot4}+\\dfrac{1}{4\\cdot6}+\\dfrac{1}{6\\cdot8}+\\cdots+ \\dfrac{1}{2014\\cdot2016}+\\dfrac{1}{2016\\cdot2018} \\, .[\/tex]<\/p>\n

\n<\/p><\/div>\n

\n

\n

Solu\u00e7\u00e3o <\/p>\n

\n

Note que, se [tex]x\\ne 0[\/tex] e [tex]x\\ne-2[\/tex]:
\n[tex]\\qquad \\qquad \\boxed{\\dfrac{1}{x\\cdot(x+2)}=\\dfrac{1}{2}\\cdot\\left( \\dfrac{1}{x} – \\dfrac{1}{x+2} \\right)} \\, .[\/tex]
\nAssim, podemos usar a igualdade acima em cada parcela da express\u00e3o a ser calculada; observe:
\n[tex]\\qquad \\dfrac{1}{2\\cdot4}=\\dfrac{1}{2}\\cdot\\left( \\dfrac{1}{2} – \\dfrac{1}{4} \\right)[\/tex]
\n[tex]\\qquad \\dfrac{1}{4\\cdot6}=\\dfrac{1}{2}\\cdot\\left( \\dfrac{1}{4} – \\dfrac{1}{6} \\right)[\/tex]
\n[tex]\\qquad \\dfrac{1}{6\\cdot8}=\\dfrac{1}{2}\\cdot\\left( \\dfrac{1}{6} – \\dfrac{1}{8} \\right)[\/tex]
\n[tex]\\qquad \\vdots[\/tex]
\n[tex]\\qquad \\dfrac{1}{2014\\cdot2016}=\\dfrac{1}{2}\\cdot\\left( \\dfrac{1}{2014} – \\dfrac{1}{2016} \\right)[\/tex]
\n[tex]\\qquad \\dfrac{1}{2016\\cdot2018}=\\dfrac{1}{2}\\cdot\\left( \\dfrac{1}{2016} – \\dfrac{1}{2018} \\right) \\, .[\/tex]
\nSomando, membro a membro, todas as parcelas acima e colocando o [tex]\\dfrac{1}{2}[\/tex] em evid\u00eancia, teremos:
\n[tex]\\qquad \\boxed{\\dfrac{1}{2\\cdot4}+\\dfrac{1}{4\\cdot6}+\\dfrac{1}{6\\cdot8}+\\cdots+ \\dfrac{1}{2014\\cdot2016}+\\dfrac{1}{2016\\cdot2018}}=[\/tex]
\n[tex]\\qquad =\\dfrac{1}{2}\\cdot\\left( \\dfrac{1}{2} – \\dfrac{1}{4} + \\dfrac{1}{4} – \\dfrac{1}{6} + \\dfrac{1}{6} – \\dfrac{1}{8} + \\cdots +\\dfrac{1}{2014} – \\dfrac{1}{2016} + \\dfrac{1}{2016} – \\dfrac{1}{2018} \\right)=[\/tex]
\n[tex]\\qquad =\\dfrac{1}{2}\\cdot\\left( \\dfrac{1}{2} – \\cancel {\\dfrac{1}{4}} + \\cancel {\\dfrac{1}{4}} – \\cancel{\\dfrac{1}{6}} +\\cancel {\\dfrac{1}{6}} -\\cancel {\\dfrac{1}{8}} + \\cdots \\\\
\n\\qquad \\qquad \\cdots + \\cancel {\\dfrac{1}{2014}} -\\cancel {\\dfrac{1}{2016}} + \\cancel{\\dfrac{1}{2016}} – \\dfrac{1}{2018} \\right)=[\/tex]
\n[tex]\\qquad =\\dfrac{1}{2}\\cdot\\left( \\dfrac{1}{2} – \\dfrac{1}{2018} \\right)=[\/tex]
\n[tex]\\qquad =\\dfrac{1}{2}\\cdot\\left( \\dfrac{1008}{2018} \\right)=[\/tex]
\n[tex]\\qquad =\\dfrac{1008}{4036}=[\/tex]
\n[tex]\\qquad = \\, \\fcolorbox{black}{#eee0e5}{$\\dfrac{252}{1009}$} \\, .[\/tex]<\/p>\n

\n

Solu\u00e7\u00e3o elaborada pelos Moderadores do Blog<\/strong>.<\/p>\n<\/div>\n

\n","protected":false},"excerpt":{"rendered":"

Problema (Indicado a partir do 9\u00ba ano do E. F.) Calcule o valor da express\u00e3o: [tex]\\dfrac{1}{2\\cdot4}+\\dfrac{1}{4\\cdot6}+\\dfrac{1}{6\\cdot8}+\\cdots+ \\dfrac{1}{2014\\cdot2016}+\\dfrac{1}{2016\\cdot2018} \\, .[\/tex] Solu\u00e7\u00e3o Note que, se [tex]x\\ne 0[\/tex] e [tex]x\\ne-2[\/tex]: [tex]\\qquad \\qquad \\boxed{\\dfrac{1}{x\\cdot(x+2)}=\\dfrac{1}{2}\\cdot\\left( \\dfrac{1}{x} – \\dfrac{1}{x+2} \\right)} \\, .[\/tex] Assim, podemos usar a igualdade acima em cada parcela da express\u00e3o a ser calculada; observe: [tex]\\qquad \\dfrac{1}{2\\cdot4}=\\dfrac{1}{2}\\cdot\\left( \\dfrac{1}{2} – … <\/p>\n

Continue lendo<\/a><\/p>\n","protected":false},"author":668,"featured_media":0,"parent":0,"menu_order":0,"comment_status":"closed","ping_status":"closed","template":"template-onecolumn.php","meta":{"footnotes":""},"class_list":["post-107024","page","type-page","status-publish","hentry","nodate","item-wrap"],"_links":{"self":[{"href":"https:\/\/clubes.obmep.org.br\/blog\/wp-json\/wp\/v2\/pages\/107024","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\/668"}],"replies":[{"embeddable":true,"href":"https:\/\/clubes.obmep.org.br\/blog\/wp-json\/wp\/v2\/comments?post=107024"}],"version-history":[{"count":9,"href":"https:\/\/clubes.obmep.org.br\/blog\/wp-json\/wp\/v2\/pages\/107024\/revisions"}],"predecessor-version":[{"id":188685,"href":"https:\/\/clubes.obmep.org.br\/blog\/wp-json\/wp\/v2\/pages\/107024\/revisions\/188685"}],"wp:attachment":[{"href":"https:\/\/clubes.obmep.org.br\/blog\/wp-json\/wp\/v2\/media?parent=107024"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}