{"id":57914,"date":"2016-03-16T19:12:26","date_gmt":"2016-03-16T22:12:26","guid":{"rendered":"http:\/\/clubes.obmep.org.br\/blog\/?page_id=57914"},"modified":"2023-03-31T20:42:22","modified_gmt":"2023-03-31T23:42:22","slug":"problema-de-olimpiada-muitas-parcelas","status":"publish","type":"page","link":"https:\/\/clubes.obmep.org.br\/blog\/problema-de-olimpiada-muitas-parcelas\/","title":{"rendered":".Problema de Olimp\u00edada: Quantas Parcelas . . ."},"content":{"rendered":"

\n

\n

Problema<\/p>\n

\nCalcule a seguinte soma:<\/p>\n

\n[tex]\\dfrac{1}{1993^{-1993}+1}+\\dfrac{1}{1993^{-1992}+1}+\\cdots+\\dfrac{1}{1993^{-1}+1}+\\dfrac{1}{1993^{0}+1}+\\\\
\n+\\dfrac{1}{1993^1+1}+\\cdots+\\dfrac{1}{1993^{1992}+1}+\\dfrac{1}{1993^{1993}+1}\\,.[\/tex]<\/p>\n

\n<\/div>\n

\n

\n

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

\nInicialmente, observe que
\n[tex]\\quad \\begin{align*}\\dfrac{1}{1993^{-x}+1}+\\dfrac{1}{1993^{x}+1}&=\\dfrac{1}{\\dfrac{1}{1993^x}+1}+\\dfrac{1}{1993^{x}+1}\\\\
\n&=\\dfrac{1}{\\dfrac{1993^x+1}{1993^x}}+\\dfrac{1}{1993^{x}+1}\\\\
\n&=\\dfrac{1993^x}{1993^x+1}+\\dfrac{1}{1993^{x}+1}\\\\
\n&=\\dfrac{1993^x+1}{1993^x+1}\\\\
\n&=1 \\,.\\end{align*}[\/tex]<\/p>\n

\n

Portanto, somando as parcelas extremas da soma, obtemos [tex]1[\/tex], o mesmo ocorrendo para as equidistantes das extremas.
\n\"oli\"<\/p>\n

Dessa forma, totalizamos [tex]1993[\/tex] parcelas iguais a [tex]1[\/tex] e mais a parcela isolada central que \u00e9 [tex]\\dfrac{1}{1993^0+1}=\\dfrac{1}{2}[\/tex].
\nObserve:<\/p>\n

\n[tex]\\dfrac{1}{1993^{-1993}+1}+\\dfrac{1}{1993^{-1992}+1}+\\cdots+\\dfrac{1}{1993^{-1}+1}+\\dfrac{1}{1993^{0}+1}+\\dfrac{1}{1993^1+1}+\\\\
\n\\quad \\cdots+\\dfrac{1}{1993^{1992}+1}+\\dfrac{1}{1993^{1993}+1}=[\/tex]
\n[tex]=\\left(\\dfrac{1}{1993^{-1993}+1}+\\dfrac{1}{1993^{1993}+1} \\right)+\\left(\\dfrac{1}{1993^{-1992}+1}+\\dfrac{1}{1993^{1992}+1} \\right)+\\cdots \\\\
\n\\quad \\cdots + \\left(\\dfrac{1}{1993^{-1}+1}+\\dfrac{1}{1993^{1}+1} \\right)+\\dfrac{1}{1993^{0}+1}= [\/tex]
\n[tex]=\\underbrace{1+1+\\cdots + 1}_{1993 \\, \\, vezes}+\\dfrac{1}{2}[\/tex].<\/p>\n

A soma final ser\u00e1, portanto, [tex]\\boxed{1993+\\dfrac{1}{2}=1993,5}[\/tex].<\/p>\n

\n

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

\n

\nOlimp\u00edada: XII ONM<\/strong> – Portugal
\n Categoria B – 10\u00ba, 11\u00ba e 12\u00ba anos <\/div>\n","protected":false},"excerpt":{"rendered":"

Problema Calcule a seguinte soma: [tex]\\dfrac{1}{1993^{-1993}+1}+\\dfrac{1}{1993^{-1992}+1}+\\cdots+\\dfrac{1}{1993^{-1}+1}+\\dfrac{1}{1993^{0}+1}+\\\\ +\\dfrac{1}{1993^1+1}+\\cdots+\\dfrac{1}{1993^{1992}+1}+\\dfrac{1}{1993^{1993}+1}\\,.[\/tex] Solu\u00e7\u00e3o Inicialmente, observe que [tex]\\quad \\begin{align*}\\dfrac{1}{1993^{-x}+1}+\\dfrac{1}{1993^{x}+1}&=\\dfrac{1}{\\dfrac{1}{1993^x}+1}+\\dfrac{1}{1993^{x}+1}\\\\ &=\\dfrac{1}{\\dfrac{1993^x+1}{1993^x}}+\\dfrac{1}{1993^{x}+1}\\\\ &=\\dfrac{1993^x}{1993^x+1}+\\dfrac{1}{1993^{x}+1}\\\\ &=\\dfrac{1993^x+1}{1993^x+1}\\\\ &=1 \\,.\\end{align*}[\/tex] Portanto, somando as parcelas extremas da soma, obtemos [tex]1[\/tex], o mesmo ocorrendo para as equidistantes das extremas. Dessa forma, totalizamos [tex]1993[\/tex] parcelas iguais a [tex]1[\/tex] e mais a parcela isolada central que \u00e9 [tex]\\dfrac{1}{1993^0+1}=\\dfrac{1}{2}[\/tex]. Observe: [tex]\\dfrac{1}{1993^{-1993}+1}+\\dfrac{1}{1993^{-1992}+1}+\\cdots+\\dfrac{1}{1993^{-1}+1}+\\dfrac{1}{1993^{0}+1}+\\dfrac{1}{1993^1+1}+\\\\ \\quad … <\/p>\n

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-57914","page","type-page","status-publish","hentry","nodate","item-wrap"],"_links":{"self":[{"href":"https:\/\/clubes.obmep.org.br\/blog\/wp-json\/wp\/v2\/pages\/57914","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=57914"}],"version-history":[{"count":33,"href":"https:\/\/clubes.obmep.org.br\/blog\/wp-json\/wp\/v2\/pages\/57914\/revisions"}],"predecessor-version":[{"id":185973,"href":"https:\/\/clubes.obmep.org.br\/blog\/wp-json\/wp\/v2\/pages\/57914\/revisions\/185973"}],"wp:attachment":[{"href":"https:\/\/clubes.obmep.org.br\/blog\/wp-json\/wp\/v2\/media?parent=57914"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}