{"id":160726,"date":"2020-08-09T19:26:32","date_gmt":"2020-08-09T22:26:32","guid":{"rendered":"http:\/\/clubes.obmep.org.br\/blog\/?page_id=160726"},"modified":"2023-07-06T05:54:04","modified_gmt":"2023-07-06T08:54:04","slug":"problema-sequencia-da-sequencia","status":"publish","type":"page","link":"https:\/\/clubes.obmep.org.br\/blog\/problema-sequencia-da-sequencia\/","title":{"rendered":".Problema: Sequ\u00eancia da sequ\u00eancia"},"content":{"rendered":"

\n

\n

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

\nCalcule o [tex]131^\\circ[\/tex] termo da sequ\u00eancia [tex]1, 2, 7, 8, 13, 14, 19, 20, 25, 26, 31, 32, 37, 38, 43, 44, 49, 50, \\dots[\/tex]<\/p>\n

Extra\u00eddo de: Manual de Progress\u00f5es<\/strong>– Lu\u00eds Lopes.<\/p>\n<\/div>\n

\n

\n

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

\nVamos denotar por [tex](a_n)[\/tex] a sequ\u00eancia dada no problema e vamos reescrev\u00ea-la destacando alguns de seus termos iniciais:
\n[tex]\\qquad \\qquad (a_n)=\\left(\\textcolor{#FF0000}{1},2,\\textcolor{#FF0000}{7},8, \\textcolor{#FF0000}{13},14,\\textcolor{#FF0000}{19},20, \\textcolor{#FF0000}{25}, 26, \\textcolor{#FF0000}{31}, 32, \\textcolor{#FF0000}{37}, 38, \\textcolor{#FF0000}{43}, 44, \\textcolor{#FF0000}{49}, 50, \\dots\\right)\\,.[\/tex]
\nObserve que os elementos em vermelho, a saber: [tex] \\textcolor{#FF0000}{1}, \\textcolor{#FF0000}{7}, \\textcolor{#FF0000}{13}, \\textcolor{#FF0000}{19}, \\textcolor{#FF0000}{25}, \\textcolor{#FF0000}{31}, \\textcolor{#FF0000}{37}, \\textcolor{#FF0000}{43}, \\textcolor{#FF0000}{49}, \\dots[\/tex] definem uma progress\u00e3o aritm\u00e9tica [tex](b_n)[\/tex] de raz\u00e3o [tex]6\\,.[\/tex]
\nPerceba, agora, que o primeiro termo da sequ\u00eancia [tex](a_n)[\/tex] corresponde ao primeiro termo da sequ\u00eancia [tex](b_n)[\/tex]; o terceiro termo da [tex](a_n)[\/tex] corresponde ao segundo termo da sequ\u00eancia [tex](b_n)[\/tex]; o quinto termo da [tex](a_n)[\/tex] corresponde ao terceiro termo da sequ\u00eancia [tex](b_n)[\/tex], e assim por diante. Vamos escrever numericamente essa correspond\u00eancia entre os termos de [tex](a_n)[\/tex] e [tex](b_n)[\/tex] para tentarmos estabelecer um padr\u00e3o:
\n[tex]\\qquad a_1=\\textcolor{#FF0000}{b_1}\\;[\/tex] e [tex]\\;\\left(\\dfrac{1-1}{2}+1 \\right)=\\textcolor{#FF0000}{1}[\/tex];
\n[tex]\\qquad a_3=\\textcolor{#FF0000}{b_2}\\;[\/tex] e [tex]\\;\\left(\\dfrac{3-1}{2}+1 \\right)=\\textcolor{#FF0000}{2}[\/tex];
\n[tex]\\qquad a_5=\\textcolor{#FF0000}{b_3}\\;[\/tex] e [tex]\\;\\left(\\dfrac{5-1}{2}+1 \\right)=\\textcolor{#FF0000}{3}[\/tex];
\n[tex]\\qquad a_7=\\textcolor{#FF0000}{b_4}\\;[\/tex] e [tex]\\;\\left(\\dfrac{7-1}{2}+1 \\right)=\\textcolor{#FF0000}{4}[\/tex].
\nCom isso, o elemento [tex]a_{131}[\/tex] corresponder\u00e1 ao termo [tex]\\textcolor{#FF0000}{\\dfrac{131-1}{2}+1=66}\\,[\/tex] de [tex]\\,(b_n)[\/tex], ou seja, [tex]\\boxed{a_{131}=\\textcolor{#FF0000}{b_{66}}}\\;.[\/tex]
\nDessa forma, para determinarmos o [tex]131^\\circ[\/tex] termo da sequ\u00eancia [tex](a_n)[\/tex], basta calcularmos o [tex]66^\\circ[\/tex] termo da sequ\u00eancia [tex](b_n)[\/tex]. Vamos l\u00e1!
\nA sequ\u00eancia [tex](b_n)[\/tex] \u00e9 uma progress\u00e3o aritm\u00e9tica cujo primeiro termo \u00e9 [tex]b_1=1[\/tex] e a raz\u00e3o \u00e9 [tex]r=6[\/tex]; portanto:
\n[tex]\\qquad b_{66}=b_1+(66-1)\\cdot r=1+65\\cdot 6=391.[\/tex]
\nPortanto, [tex]\\fcolorbox{black}{#eee0e5}{$a_{131}=b_{66}=391$}\\;.[\/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.) Calcule o [tex]131^\\circ[\/tex] termo da sequ\u00eancia [tex]1, 2, 7, 8, 13, 14, 19, 20, 25, 26, 31, 32, 37, 38, 43, 44, 49, 50, \\dots[\/tex] Extra\u00eddo de: Manual de Progress\u00f5es– Lu\u00eds Lopes. Solu\u00e7\u00e3o Vamos denotar por [tex](a_n)[\/tex] a sequ\u00eancia dada no problema e vamos reescrev\u00ea-la … <\/p>\n

Continue lendo<\/a><\/p>\n","protected":false},"author":164,"featured_media":0,"parent":0,"menu_order":0,"comment_status":"closed","ping_status":"closed","template":"template-onecolumn.php","meta":{"footnotes":""},"class_list":["post-160726","page","type-page","status-publish","hentry","nodate","item-wrap"],"_links":{"self":[{"href":"https:\/\/clubes.obmep.org.br\/blog\/wp-json\/wp\/v2\/pages\/160726","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\/164"}],"replies":[{"embeddable":true,"href":"https:\/\/clubes.obmep.org.br\/blog\/wp-json\/wp\/v2\/comments?post=160726"}],"version-history":[{"count":37,"href":"https:\/\/clubes.obmep.org.br\/blog\/wp-json\/wp\/v2\/pages\/160726\/revisions"}],"predecessor-version":[{"id":196965,"href":"https:\/\/clubes.obmep.org.br\/blog\/wp-json\/wp\/v2\/pages\/160726\/revisions\/196965"}],"wp:attachment":[{"href":"https:\/\/clubes.obmep.org.br\/blog\/wp-json\/wp\/v2\/media?parent=160726"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}