{"id":11887,"date":"2014-05-22T16:20:37","date_gmt":"2014-05-22T15:20:37","guid":{"rendered":"https:\/\/www.acasinhadamatematica.pt\/?p=11887"},"modified":"2022-01-21T02:48:26","modified_gmt":"2022-01-21T02:48:26","slug":"mostre-que-e-limitada-a-sucessao-de-termo-geral-u_n-left-frac15-rightnfrac4nleft-1-rightn-85n-3","status":"publish","type":"post","link":"https:\/\/www.acasinhadamatematica.pt\/?p=11887","title":{"rendered":"Mostre que \u00e9 limitada a sucess\u00e3o de termo geral ${u_n} = {\\left( { &#8211; \\frac{1}{5}} \\right)^n}\\frac{{4n{{\\left( { &#8211; 1} \\right)}^n} &#8211; 8}}{{5n + 3}}$"},"content":{"rendered":"<p><ul id='GTTabs_ul_11887' class='GTTabs' style='display:none'>\n<li id='GTTabs_li_0_11887' class='GTTabs_curr'><a  id=\"11887_0\" onMouseOver=\"GTTabsShowLinks('Enunciado'); return true;\"  onMouseOut=\"GTTabsShowLinks();\"  class='GTTabsLinks'>Enunciado<\/a><\/li>\n<li id='GTTabs_li_1_11887' ><a  id=\"11887_1\" onMouseOver=\"GTTabsShowLinks('Resolu\u00e7\u00e3o'); return true;\"  onMouseOut=\"GTTabsShowLinks();\"  class='GTTabsLinks'>Resolu\u00e7\u00e3o<\/a><\/li>\n<\/ul>\n\n<div class='GTTabs_divs GTTabs_curr_div' id='GTTabs_0_11887'>\n<span class='GTTabs_titles'><b>Enunciado<\/b><\/span><\/p>\n<p>Mostre que \u00e9 limitada a sucess\u00e3o de termo geral ${u_n} = {\\left( { &#8211; \\frac{1}{5}} \\right)^n}\\frac{{4n{{\\left( { &#8211; 1} \\right)}^n} &#8211; 8}}{{5n + 3}}$.<\/p>\n<p><div class='GTTabsNavigation' style='display:none'><span class='GTTabs_nav_next'><a href='#GTTabs_ul_11887' onClick='GTTabs_show(1,11887)'>Resolu\u00e7\u00e3o &gt;&gt;<\/a><\/span><\/div><\/div>\n\n<div class='GTTabs_divs' id='GTTabs_1_11887'>\n<span class='GTTabs_titles'><b>Resolu\u00e7\u00e3o<\/b><\/span><!--more--><\/p>\n<blockquote>\n<p>Mostre que \u00e9 limitada a sucess\u00e3o de termo geral ${u_n} = {\\left( { &#8211; \\frac{1}{5}} \\right)^n}\\frac{{4n{{\\left( { &#8211; 1} \\right)}^n} &#8211; 8}}{{5n + 3}}$.<\/p>\n<\/blockquote>\n<p>\u00ad<\/p>\n<p>Consideremos duas sucess\u00f5es, \\(\\left( {{a_n}} \\right)\\) e \\(\\left( {{b_n}} \\right)\\), tais que os termos de ordem \u00edmpar de \\(\\left( {{u_n}} \\right)\\) s\u00e3o termos da primeira e os termos de ordem par de \\(\\left( {{u_n}} \\right)\\) s\u00e3o termos da segunda:<\/p>\n<table class=\" aligncenter\" style=\"width: 60%;\">\n<tbody>\n<tr>\n<td>\n<p>\\(n\\)<\/p>\n<\/td>\n<td width=\"71\">\n<p>\\(1\\)<\/p>\n<\/td>\n<td width=\"71\">\n<p>\\(2\\)<\/p>\n<\/td>\n<td width=\"71\">\n<p>\\(3\\)<\/p>\n<\/td>\n<td width=\"71\">\n<p>\\(4\\)<\/p>\n<\/td>\n<td width=\"71\">\n<p>\\(5\\)<\/p>\n<\/td>\n<td width=\"71\">\n<p>\u2026<\/p>\n<\/td>\n<\/tr>\n<tr>\n<td>\n<p>\\({a_n}\\)<\/p>\n<\/td>\n<td width=\"71\">\n<p><span style=\"color: #ff0000;\">\\(\\frac{3}{{10}}\\)<\/span><\/p>\n<\/td>\n<td width=\"71\">\n<p><span style=\"color: #ff0000;\">\\(\\frac{{16}}{{325}}\\)<\/span><\/p>\n<\/td>\n<td width=\"71\">\n<p><span style=\"color: #ff0000;\">\\(\\frac{2}{{225}}\\)<\/span><\/p>\n<\/td>\n<td width=\"71\">\n<p><span style=\"color: #ff0000;\">\\(\\frac{{24}}{{14375}}\\)<\/span><\/p>\n<\/td>\n<td width=\"71\">\n<p><span style=\"color: #ff0000;\">\\(\\frac{1}{{3125}}\\)<\/span><\/p>\n<\/td>\n<td width=\"71\">\n<p>\u2026<\/p>\n<\/td>\n<\/tr>\n<tr>\n<td>\n<p>\\({u_n}\\)<\/p>\n<\/td>\n<td width=\"71\">\n<p><span style=\"color: #ff0000;\">\\(\\frac{3}{{10}}\\)<\/span><\/p>\n<\/td>\n<td width=\"71\">\n<p><span style=\"color: #008000;\">\\(0\\)<\/span><\/p>\n<\/td>\n<td width=\"71\">\n<p><span style=\"color: #ff0000;\">\\(\\frac{2}{{225}}\\)<\/span><\/p>\n<\/td>\n<td width=\"71\">\n<p><span style=\"color: #008000;\">\\(\\frac{8}{{14375}}\\)<\/span><\/p>\n<\/td>\n<td width=\"71\">\n<p><span style=\"color: #ff0000;\">\\(\\frac{1}{{3125}}\\)<\/span><\/p>\n<\/td>\n<td width=\"71\">\n<p>\u2026<\/p>\n<\/td>\n<\/tr>\n<tr>\n<td>\n<p>\\({b_n}\\)<\/p>\n<\/td>\n<td width=\"71\">\n<p><span style=\"color: #008000;\">\\( &#8211; \\frac{1}{{10}}\\)<\/span><\/p>\n<\/td>\n<td width=\"71\">\n<p><span style=\"color: #008000;\">\\(0\\)<\/span><\/p>\n<\/td>\n<td width=\"71\">\n<p><span style=\"color: #008000;\">\\(\\frac{2}{{1125}}\\)<\/span><\/p>\n<\/td>\n<td width=\"71\">\n<p><span style=\"color: #008000;\">\\(\\frac{8}{{14375}}\\)<\/span><\/p>\n<\/td>\n<td width=\"71\">\n<p><span style=\"color: #008000;\">\\(\\frac{3}{{21875}}\\)<\/span><\/p>\n<\/td>\n<td width=\"71\">\n<p>\u2026<\/p>\n<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<p>Desta forma, caso as sucess\u00f5es \\(\\left( {{a_n}} \\right)\\) e \\(\\left( {{b_n}} \\right)\\) sejam limitadas, ent\u00e3o tamb\u00e9m ser\u00e1 limitada a sucess\u00e3o \\(\\left( {{u_n}} \\right)\\).<\/p>\n<\/p>\n<p>Essas sucess\u00f5es podem ser assim definidas:<\/p>\n<ul>\n<li>\\({a_n} =\u00a0 &#8211; {\\left( {\\frac{1}{5}} \\right)^n}\\frac{{ &#8211; 4n &#8211; 8}}{{5n + 3}} = \\frac{1}{{{5^n}}} \\times \\frac{{4n + 8}}{{5n + 3}} = \\frac{1}{{{5^n}}} \\times \\left( {\\frac{4}{5} + \\frac{{{\\textstyle{{28} \\over 5}}}}{{5n + 3}}} \\right)\\)<\/li>\n<li>\\({b_n} = {\\left( {\\frac{1}{5}} \\right)^n}\\frac{{4n &#8211; 8}}{{5n + 3}} = \\frac{1}{{{5^n}}} \\times \\frac{{4n &#8211; 8}}{{5n + 3}} = \\frac{1}{{{5^n}}} \\times \\left( {\\frac{4}{5} &#8211; \\frac{{{\\textstyle{{52} \\over 5}}}}{{5n + 3}}} \\right)\\)<\/li>\n<\/ul>\n<p>Ora, a sucess\u00e3o \\(\\left( {{a_n}} \\right)\\) \u00e9 estritamente decrescente, pois \u00e9 o produto de duas sucess\u00f5es estritamente decrescentes: \\({c_n} = \\frac{1}{{{5^n}}}\\) e \\({d_n} = \\frac{4}{5} + \\frac{{{\\textstyle{{28} \\over 5}}}}{{5n + 3}}\\).<\/p>\n<p>Como todos os termos da sucess\u00e3o \\(\\left( {{a_n}} \\right)\\) s\u00e3o positivos e \\({a_1} = \\frac{{12}}{{40}} = \\frac{3}{{10}}\\), conclui-se que esta sucess\u00e3o \u00e9 limitada, pois tem-se: $\\boxed{0 &lt; {a_n} \\leqslant \\frac{3}{{10}},\\forall n \\in \\mathbb{N}}$.<\/p>\n<\/p>\n<p>Passemos a estudar a monotonia da sucess\u00e3o \\(\\left( {{b_n}} \\right)\\):<\/p>\n<p>\\[\\begin{array}{*{20}{l}}<br \/>\n{{b_{n + 1}} &#8211; {b_n}}&amp; = &amp;{\\frac{1}{{{5^{n + 1}}}} \\times \\frac{{4\\left( {n + 1} \\right) &#8211; 8}}{{5\\left( {n + 1} \\right) + 3}} &#8211; \\frac{1}{{{5^n}}} \\times \\frac{{4n &#8211; 8}}{{5n + 3}}} \\\\<br \/>\n{}&amp; = &amp;{\\frac{1}{{{5^{n + 1}}}} \\times \\frac{{4n &#8211; 4}}{{5n + 8}} &#8211; \\frac{1}{{{5^n}}} \\times \\frac{{4n &#8211; 8}}{{5n + 3}}} \\\\<br \/>\n{}&amp; = &amp;{\\frac{1}{{{5^n}}}\\left( {\\frac{{4n &#8211; 4}}{{25n + 40}} &#8211; \\frac{{4n &#8211; 8}}{{5n + 3}}} \\right)} \\\\<br \/>\n{}&amp; = &amp;{\\frac{1}{{{5^n}}} \\times \\frac{{20{n^2} + 12n &#8211; 20n &#8211; 12 &#8211; 100{n^2} + 200n &#8211; 160n + 320}}{{\\left( {25n + 40} \\right)\\left( {5n + 3} \\right)}}} \\\\<br \/>\n{}&amp; = &amp;{\\frac{1}{{{5^n}}} \\times \\frac{{ &#8211; 80{n^2} + 32n + 308}}{{\\left( {25n + 40} \\right)\\left( {5n + 3} \\right)}}} \\\\<br \/>\n{}&amp; = &amp;{\\frac{4}{{{5^n}}} \\times \\frac{{ &#8211; 20{n^2} + 8n + 77}}{{\\left( {25n + 40} \\right)\\left( {5n + 3} \\right)}}}<br \/>\n\\end{array}\\]<\/p>\n<\/p>\n<p><a href=\"https:\/\/www.acasinhadamatematica.pt\/wp-content\/uploads\/2014\/05\/Px.png\"><img loading=\"lazy\" decoding=\"async\" data-attachment-id=\"11888\" data-permalink=\"https:\/\/www.acasinhadamatematica.pt\/?attachment_id=11888\" data-orig-file=\"https:\/\/www.acasinhadamatematica.pt\/wp-content\/uploads\/2014\/05\/Px.png\" data-orig-size=\"513,514\" data-comments-opened=\"1\" data-image-meta=\"{&quot;aperture&quot;:&quot;0&quot;,&quot;credit&quot;:&quot;&quot;,&quot;camera&quot;:&quot;&quot;,&quot;caption&quot;:&quot;&quot;,&quot;created_timestamp&quot;:&quot;0&quot;,&quot;copyright&quot;:&quot;&quot;,&quot;focal_length&quot;:&quot;0&quot;,&quot;iso&quot;:&quot;0&quot;,&quot;shutter_speed&quot;:&quot;0&quot;,&quot;title&quot;:&quot;&quot;}\" data-image-title=\"Gr\u00e1fico de P\" data-image-description=\"\" data-image-caption=\"\" data-large-file=\"https:\/\/www.acasinhadamatematica.pt\/wp-content\/uploads\/2014\/05\/Px.png\" class=\"alignright wp-image-11888\" src=\"https:\/\/www.acasinhadamatematica.pt\/wp-content\/uploads\/2014\/05\/Px.png\" alt=\"Gr\u00e1fico de P\" width=\"250\" height=\"250\" srcset=\"https:\/\/www.acasinhadamatematica.pt\/wp-content\/uploads\/2014\/05\/Px.png 513w, https:\/\/www.acasinhadamatematica.pt\/wp-content\/uploads\/2014\/05\/Px-150x150.png 150w, https:\/\/www.acasinhadamatematica.pt\/wp-content\/uploads\/2014\/05\/Px-300x300.png 300w, https:\/\/www.acasinhadamatematica.pt\/wp-content\/uploads\/2014\/05\/Px-400x400.png 400w\" sizes=\"auto, (max-width: 250px) 100vw, 250px\" \/><\/a>Determinemos os zeros do polin\u00f3mio \\(P\\left( x \\right) =\u00a0 &#8211; 20{x^2} + 8x + 77\\):<\/p>\n<p>\\[\\begin{array}{*{20}{l}}<br \/>\n{P\\left( x \\right) = 0}&amp; \\Leftrightarrow &amp;{x = \\frac{{ &#8211; 8 \\pm \\sqrt {{8^2} + 80 \\times 77} }}{{ &#8211; 40}}} \\\\<br \/>\n{}&amp; \\Leftrightarrow &amp;{x = \\frac{{ &#8211; 8 \\pm \\sqrt {6224} }}{{ &#8211; 40}}} \\\\<br \/>\n{}&amp; \\Leftrightarrow &amp;{x = \\frac{{ &#8211; 8 \\pm 4\\sqrt {389} }}{{ &#8211; 40}}} \\\\<br \/>\n{}&amp; \\Leftrightarrow &amp;{x = \\frac{1}{5} \\pm \\frac{{\\sqrt {389} }}{{10}}} \\\\<br \/>\n{}&amp;{}&amp;{\\begin{array}{*{20}{l}}<br \/>\n{{x_1} \\approx\u00a0 &#8211; 1,77}&amp; \\vee &amp;{{x_2} \\approx 2,17}<br \/>\n\\end{array}}<br \/>\n\\end{array}\\]<\/p>\n<\/p>\n<p>Face aos valores dos zeros de \\(P\\left( x \\right) =\u00a0 &#8211; 20{x^2} + 8x + 77\\), podemos concluir:<\/p>\n<p>\\[{b_{n + 1}} &#8211; {b_n} &lt; 0,\\forall n \\in \\mathbb{N}\\backslash \\left\\{ {1,2} \\right\\}\\]<\/p>\n<p>Ou seja, ainda que a sucess\u00e3o \\(\\left( {{b_n}} \\right)\\) n\u00e3o seja mon\u00f3tona, ela \u00e9 uma fun\u00e7\u00e3o decrescente para \\(n \\in \\mathbb{N}\\backslash \\left\\{ {1,2} \\right\\}\\).<\/p>\n<\/p>\n<p>Considerando o resultado anterior, considerando que a sucess\u00e3o \\(\\left( {{b_n}} \\right)\\) n\u00e3o possuiu qualquer termo negativo e que ${b_3} = \\frac{2}{{1125}}$, podemos concluir: $\\boxed{0 &lt; {b_n} \\leqslant \\frac{2}{{1125}},\\forall n \\in \\mathbb{N}\\backslash \\left\\{ {1,2} \\right\\}}$.<\/p>\n<p>Dado que\u00a0${b_1} =\u00a0 &#8211; \\frac{1}{{10}}$ e ${b_2} = 0$, conclui-se que \\(\\left( {{b_n}} \\right)\\) \u00e9 tamb\u00e9m uma sucess\u00e3o limitada, pois tem-se: $\\boxed{ &#8211; \\frac{1}{{10}} \\leqslant {b_n} \\leqslant \\frac{2}{{1125}},\\forall n \\in \\mathbb{N}}$.<\/p>\n<\/p>\n<p>Assim, como as sucess\u00f5es \\(\\left( {{a_n}} \\right)\\) e \\(\\left( {{b_n}} \\right)\\) s\u00e3o limitadas, $\\boxed{0 &lt; {a_n} \\leqslant \\frac{3}{{10}},\\forall n \\in \\mathbb{N}}$\u00a0e $\\boxed{ &#8211; \\frac{1}{{10}} \\leqslant {b_n} \\leqslant \\frac{2}{{1125}},\\forall n \\in \\mathbb{N}}$, ent\u00e3o tamb\u00e9m \u00e9 limitada a sucess\u00e3o \\(\\left( {{u_n}} \\right)\\), n\u00e3o mon\u00f3tona, pois tem-se: $\\boxed{ &#8211; \\frac{1}{{10}} \\leqslant {u_n} \\leqslant \\frac{3}{{10}},\\forall n \\in \\mathbb{N}}$.<\/p>\n<\/p>\n<p>No entanto, n\u00e3o esquecendo que nem todos os termos de \\(\\left( {{a_n}} \\right)\\) e de \\(\\left( {{b_n}} \\right)\\) s\u00e3o termos da sucess\u00e3o \\(\\left( {{u_n}} \\right)\\) e considerando o que j\u00e1 foi exposto anteriormente, podemos ainda restringir o enquadramento dos termos da sucess\u00e3o \\(\\left( {{u_n}} \\right)\\): $\\boxed{0 \\leqslant {u_n} \\leqslant \\frac{3}{{10}},\\forall n \\in \\mathbb{N}}$.<\/p>\n<\/p>\n<p>Apresenta-se, de seguida, a representa\u00e7\u00e3o gr\u00e1fica dos primeiros 10 termos de \\(\\left( {{u_n}} \\right)\\).<\/p>\n<p>\u00a0<a href=\"https:\/\/www.acasinhadamatematica.pt\/wp-content\/uploads\/2014\/05\/Termos.png\"><img loading=\"lazy\" decoding=\"async\" data-attachment-id=\"11889\" data-permalink=\"https:\/\/www.acasinhadamatematica.pt\/?attachment_id=11889\" data-orig-file=\"https:\/\/www.acasinhadamatematica.pt\/wp-content\/uploads\/2014\/05\/Termos.png\" data-orig-size=\"791,451\" data-comments-opened=\"1\" data-image-meta=\"{&quot;aperture&quot;:&quot;0&quot;,&quot;credit&quot;:&quot;&quot;,&quot;camera&quot;:&quot;&quot;,&quot;caption&quot;:&quot;&quot;,&quot;created_timestamp&quot;:&quot;0&quot;,&quot;copyright&quot;:&quot;&quot;,&quot;focal_length&quot;:&quot;0&quot;,&quot;iso&quot;:&quot;0&quot;,&quot;shutter_speed&quot;:&quot;0&quot;,&quot;title&quot;:&quot;&quot;}\" data-image-title=\"Primeiros termos da sucess\u00e3o\" data-image-description=\"\" data-image-caption=\"\" data-large-file=\"https:\/\/www.acasinhadamatematica.pt\/wp-content\/uploads\/2014\/05\/Termos.png\" class=\"aligncenter wp-image-11889\" src=\"https:\/\/www.acasinhadamatematica.pt\/wp-content\/uploads\/2014\/05\/Termos.png\" alt=\"Primeiros termos da sucess\u00e3o\" width=\"600\" height=\"342\" srcset=\"https:\/\/www.acasinhadamatematica.pt\/wp-content\/uploads\/2014\/05\/Termos.png 791w, https:\/\/www.acasinhadamatematica.pt\/wp-content\/uploads\/2014\/05\/Termos-300x171.png 300w, https:\/\/www.acasinhadamatematica.pt\/wp-content\/uploads\/2014\/05\/Termos-150x85.png 150w, https:\/\/www.acasinhadamatematica.pt\/wp-content\/uploads\/2014\/05\/Termos-400x228.png 400w\" sizes=\"auto, (max-width: 600px) 100vw, 600px\" \/><\/a><\/p><\/p>\n<div class='GTTabsNavigation' style='display:none'><span class='GTTabs_nav_prev'><a href='#GTTabs_ul_11887' onClick='GTTabs_show(0,11887)'>&lt;&lt; Enunciado<\/a><\/span><\/div><\/div>\n\n","protected":false},"excerpt":{"rendered":"<p>Enunciado Resolu\u00e7\u00e3o Enunciado Mostre que \u00e9 limitada a sucess\u00e3o de termo geral ${u_n} = {\\left( { &#8211; \\frac{1}{5}} \\right)^n}\\frac{{4n{{\\left( { &#8211; 1} \\right)}^n} &#8211; 8}}{{5n + 3}}$. Resolu\u00e7\u00e3o &gt;&gt; Resolu\u00e7\u00e3o &lt;&lt; Enunciado<\/p>\n","protected":false},"author":1,"featured_media":20809,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"_jetpack_memberships_contains_paid_content":false,"footnotes":""},"categories":[98,97,372],"tags":[422,373,431],"series":[],"class_list":["post-11887","post","type-post","status-publish","format-standard","has-post-thumbnail","hentry","category-11--ano","category-aplicando","category-sucessoes-reais","tag-11-o-ano","tag-sucessao-limitada","tag-sucessoes-reais"],"views":3344,"jetpack_featured_media_url":"https:\/\/www.acasinhadamatematica.pt\/wp-content\/uploads\/2014\/05\/11V3Pag057-11_520x245.png","jetpack_sharing_enabled":true,"jetpack_likes_enabled":true,"_links":{"self":[{"href":"https:\/\/www.acasinhadamatematica.pt\/index.php?rest_route=\/wp\/v2\/posts\/11887","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/www.acasinhadamatematica.pt\/index.php?rest_route=\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/www.acasinhadamatematica.pt\/index.php?rest_route=\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/www.acasinhadamatematica.pt\/index.php?rest_route=\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/www.acasinhadamatematica.pt\/index.php?rest_route=%2Fwp%2Fv2%2Fcomments&post=11887"}],"version-history":[{"count":0,"href":"https:\/\/www.acasinhadamatematica.pt\/index.php?rest_route=\/wp\/v2\/posts\/11887\/revisions"}],"wp:featuredmedia":[{"embeddable":true,"href":"https:\/\/www.acasinhadamatematica.pt\/index.php?rest_route=\/wp\/v2\/media\/20809"}],"wp:attachment":[{"href":"https:\/\/www.acasinhadamatematica.pt\/index.php?rest_route=%2Fwp%2Fv2%2Fmedia&parent=11887"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.acasinhadamatematica.pt\/index.php?rest_route=%2Fwp%2Fv2%2Fcategories&post=11887"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.acasinhadamatematica.pt\/index.php?rest_route=%2Fwp%2Fv2%2Ftags&post=11887"},{"taxonomy":"series","embeddable":true,"href":"https:\/\/www.acasinhadamatematica.pt\/index.php?rest_route=%2Fwp%2Fv2%2Fseries&post=11887"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}