{"id":160771,"date":"2020-08-09T22:47:52","date_gmt":"2020-08-10T01:47:52","guid":{"rendered":"http:\/\/clubes.obmep.org.br\/blog\/?page_id=160771"},"modified":"2023-07-06T05:53:33","modified_gmt":"2023-07-06T08:53:33","slug":"problema-uma-complicada-soma-de-fracoes","status":"publish","type":"page","link":"https:\/\/clubes.obmep.org.br\/blog\/problema-uma-complicada-soma-de-fracoes\/","title":{"rendered":".Problema: Uma complicada soma de fra\u00e7\u00f5es"},"content":{"rendered":"

\n

\n

Problema
\n (Indicado a partir do 2\u00ba ano do E. M.)<\/span><\/p>\n

\nMostre a validade da seguinte igualdade, para [tex]n\\geq 3[\/tex].
\n
[tex]\\dfrac{1}{1 \\cdot 3 \\cdot 5}+\\dfrac{1}{3 \\cdot 5 \\cdot 7}+\\dfrac{1}{5 \\cdot 7 \\cdot 9}+\\dots +\\dfrac{1}{(n-2) \\cdot n \\cdot (n+2)}=\\dfrac{1} {4} \\left(\\dfrac{1}{3}-\\dfrac {1}{n \\cdot(n+2)}\\right).[\/tex]<\/center>\n<\/div>\n

\n

\n

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

\nInicialmente, vamos provar que, para todo [tex]k[\/tex] real, com [tex]k \\neq -2,\\,0,\\, 2[\/tex], vale a seguinte igualdade:<\/li>\n

[tex] \\dfrac{1}{(k-2)k(k+2)} = \\dfrac{1}{4} \\left( \\dfrac{1}{(k-2)k}-\\dfrac{1}{k(k+2)} \\right). \\qquad \\textcolor{#800000}{(i)}[\/tex] <\/p>\n

Com efeito:
\n[tex]\\qquad \\begin{aligned}\\dfrac{1} {4} \\cdot \\left(\\dfrac {1}{(k-2) \\cdot k}-\\dfrac{1}{k\\cdot (k+2)} \\right) &= \\dfrac{1}{4} \\left( \\dfrac{(k+2)-(k-2)}{(k-2)k(k+2)} \\right)\\\\
\n &= \\dfrac{1}{4}\\left( \\dfrac{4}{(k-2)k(k+2)} \\right)\\\\
\n &= \\dfrac{1}{(k-2)k(k+2)},\\end{aligned}[\/tex]
\n como quer\u00edamos mostrar.
\nNote que, segue de [tex]\\textcolor{#800000}{(i)}[\/tex], que:
\n[tex]\\qquad\\dfrac{1}{1 \\cdot 3 \\cdot 5}= \\dfrac{1} {4} \\left( \\dfrac {1}{1 \\cdot 3} – \\dfrac {1}{3 \\cdot 5} \\right)[\/tex];<\/p>\n

\n

[tex]\\qquad\\dfrac{1}{3 \\cdot 5 \\cdot 7} = \\dfrac{1} {4} \\left( \\dfrac {1}{3 \\cdot 5} – \\dfrac {1}{5 \\cdot 7} \\right)[\/tex];<\/p>\n

\n

[tex]\\qquad\\dfrac{1}{5 \\cdot 7 \\cdot 9} = \\dfrac{1} {4} \\left( \\dfrac {1}{5 \\cdot 7} – \\dfrac {1}{7 \\cdot 9} \\right)[\/tex];
\n[tex]\\qquad\\;\\; \\vdots\\\\[\/tex]
\n[tex]\\qquad\\dfrac{1}{(n-2) \\cdot n \\cdot (n-2)}=\\dfrac{1} {4} \\cdot \\left(\\dfrac {1}{(n-2) \\cdot n} – \\dfrac {1}{n \\cdot (n+2)}\\right)[\/tex].<\/p>\n

\n

Somando-se todas as igualdades acima, obtemos:<\/p>\n

[tex]\\qquad\\dfrac{1}{1 \\cdot 3 \\cdot 5} + \\dfrac{1}{3 \\cdot 5 \\cdot 7} + \\dfrac{1}{5 \\cdot 7 \\cdot 9} + \\dots + \\dfrac{1}{ (n-2) \\cdot n \\cdot (n+2)} = \\\\
\n\\qquad=\\dfrac{1} {4} \\cdot\\left[\\left( \\dfrac {1}{1 \\cdot 3} – \\textcolor{red}{\\cancel{\\dfrac {1}{3 \\cdot 5}}} \\right)+\\left( \\textcolor{red}{ \\cancel{\\dfrac {1}{3 \\cdot 5}}} – \\textcolor{blue}{\\bcancel{\\dfrac {1}{5 \\cdot 7}}} \\right)+\\left( \\textcolor{blue}{\\bcancel{ \\dfrac {1}{5 \\cdot 7}}} – \\textcolor{green}{\\cancel{\\dfrac {1}{7 \\cdot 9}}} \\right)+\\cdots\\\\
\n\\qquad \\quad \\cdots +\\left(\\textcolor{#FF00FF}{\\bcancel{\\dfrac {1}{(n-2) \\cdot n}}} – \\dfrac {1}{n \\cdot (n+2)}\\right) \\right]\\\\
\n\\qquad = \\dfrac{1} {4} \\left( \\dfrac {1}{ 1 \\cdot 3} – \\dfrac {1}{n \\cdot (n+2)} \\right)\\,.[\/tex]
\nPortanto, de fato,<\/p>\n

[tex]\\fcolorbox{black}{#eee0e5}{$\\dfrac{1}{1 \\cdot 3 \\cdot 5} + \\dfrac{1}{3 \\cdot 5 \\cdot 7} + \\dfrac{1}{5 \\cdot 7 \\cdot 9} + \\dots + \\dfrac{1}{ (n-2) \\cdot n \\cdot (n+2)} = \\dfrac{1} {4} \\left( \\dfrac {1}{3} – \\dfrac {1}{n \\cdot (n+2)} \\right)$}\\,.[\/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 2\u00ba ano do E. M.) Mostre a validade da seguinte igualdade, para [tex]n\\geq 3[\/tex]. [tex]\\dfrac{1}{1 \\cdot 3 \\cdot 5}+\\dfrac{1}{3 \\cdot 5 \\cdot 7}+\\dfrac{1}{5 \\cdot 7 \\cdot 9}+\\dots +\\dfrac{1}{(n-2) \\cdot n \\cdot (n+2)}=\\dfrac{1} {4} \\left(\\dfrac{1}{3}-\\dfrac {1}{n \\cdot(n+2)}\\right).[\/tex] Solu\u00e7\u00e3o Inicialmente, vamos provar que, para todo [tex]k[\/tex] real, com [tex]k \\neq -2,\\,0,\\, 2[\/tex], … <\/p>\n

Continue lendo<\/a><\/p>\n","protected":false},"author":732,"featured_media":0,"parent":0,"menu_order":0,"comment_status":"closed","ping_status":"closed","template":"template-onecolumn.php","meta":{"footnotes":""},"class_list":["post-160771","page","type-page","status-publish","hentry","nodate","item-wrap"],"_links":{"self":[{"href":"https:\/\/clubes.obmep.org.br\/blog\/wp-json\/wp\/v2\/pages\/160771","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\/732"}],"replies":[{"embeddable":true,"href":"https:\/\/clubes.obmep.org.br\/blog\/wp-json\/wp\/v2\/comments?post=160771"}],"version-history":[{"count":53,"href":"https:\/\/clubes.obmep.org.br\/blog\/wp-json\/wp\/v2\/pages\/160771\/revisions"}],"predecessor-version":[{"id":196964,"href":"https:\/\/clubes.obmep.org.br\/blog\/wp-json\/wp\/v2\/pages\/160771\/revisions\/196964"}],"wp:attachment":[{"href":"https:\/\/clubes.obmep.org.br\/blog\/wp-json\/wp\/v2\/media?parent=160771"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}