{"id":779,"date":"2021-09-22T05:00:14","date_gmt":"2021-09-22T03:00:14","guid":{"rendered":"https:\/\/www.glc.us.es\/~jalonso\/calculemus\/?p=779"},"modified":"2021-09-17T16:49:46","modified_gmt":"2021-09-17T14:49:46","slug":"suma-de-los-primeros-cubos","status":"publish","type":"post","link":"https:\/\/www.glc.us.es\/~jalonso\/calculemus\/suma-de-los-primeros-cubos\/","title":{"rendered":"Suma de los primeros cubos"},"content":{"rendered":"<p>Demostrar que la suma de los primeros cubos<\/p>\n<pre lang=\"text\">\n   0\u00b3 + 1\u00b3 + 2\u00b3 + 3\u00b3 + \u00b7\u00b7\u00b7 + n\u00b3\n<\/pre>\n<p>es (n(n+1)\/2)\u00b2<\/p>\n<p>Para ello, completar la siguiente teor\u00eda de Lean:<\/p>\n<pre lang=\"lean\">\nimport data.nat.basic\nimport tactic\nopen nat\n\nvariable (n : \u2115)\n\ndef sumaCubos : \u2115 \u2192 \u2115\n| 0     := 0\n| (n+1) := sumaCubos n + (n+1)^3\n\nexample :\n  4 * sumaCubos n = (n*(n+1))^2 :=\nsorry\n<\/pre>\n<p>[expand title=\u00bbSoluciones con Lean\u00bb]<\/p>\n<pre lang=\"lean\">\r\nimport data.nat.basic\r\nimport tactic\r\nopen nat\r\n\r\nvariable (n : \u2115)\r\n\r\nset_option pp.structure_projections false\r\n\r\n@[simp]\r\ndef sumaCubos : \u2115 \u2192 \u2115\r\n| 0     := 0\r\n| (n+1) := sumaCubos n + (n+1)^3\r\n\r\n-- 1\u00aa demostraci\u00f3n\r\nexample :\r\n  4 * sumaCubos n = (n*(n+1))^2 :=\r\nbegin\r\n  induction n with n HI,\r\n  { simp,\r\n    ring, },\r\n  { calc 4 * sumaCubos (succ n)\r\n         = 4 * (sumaCubos n + (n+1)^3)\r\n           : by simp\r\n     ... = 4 * sumaCubos n + 4*(n+1)^3\r\n           : by ring\r\n     ... = (n*(n+1))^2 + 4*(n+1)^3\r\n           : by {congr; rw HI}\r\n     ... = (n+1)^2*(n^2+4*n+4)\r\n           : by ring\r\n     ... = (n+1)^2*(n+2)^2\r\n           : by ring\r\n     ... = ((n+1)*(n+2))^2\r\n           : by ring\r\n     ... = (succ n * (succ n + 1)) ^ 2\r\n           : by simp, },\r\nend\r\n\r\n-- 2\u00aa demostraci\u00f3n\r\nexample :\r\n  4 * sumaCubos n = (n*(n+1))^2 :=\r\nbegin\r\n  induction n with n HI,\r\n  { simp,\r\n    ring, },\r\n  { calc 4 * sumaCubos (succ n)\r\n         = 4 * sumaCubos n + 4*(n+1)^3\r\n           : by {simp ; ring}\r\n     ... = (n*(n+1))^2 + 4*(n+1)^3\r\n           : by {congr; rw HI}\r\n     ... = ((n+1)*(n+2))^2\r\n           : by ring\r\n     ... = (succ n * (succ n + 1)) ^ 2\r\n           : by simp, },\r\nend\r\n<\/pre>\n<p>Se puede interactuar con la prueba anterior en <a href=\"https:\/\/leanprover-community.github.io\/lean-web-editor\/#url=https:\/\/raw.githubusercontent.com\/jaalonso\/Calculemus\/main\/src\/Suma_de_los_primeros_cubos.lean\" rel=\"noopener noreferrer\" target=\"_blank\">esta sesi\u00f3n con Lean<\/a>.<\/p>\n<p>En los comentarios se pueden escribir otras soluciones, escribiendo el c\u00f3digo entre una l\u00ednea con &#60;pre lang=&quot;lean&quot;&#62; y otra con &#60;\/pre&#62;<br \/>\n[\/expand]<\/p>\n<p>[expand title=\u00bbSoluciones con Isabelle\/HOL\u00bb]<\/p>\n<pre lang=\"isar\">\r\ntheory Suma_de_los_primeros_cubos\r\nimports Main\r\nbegin\r\n\r\nfun sumaCubos :: \"nat \u21d2 nat\" where\r\n  \"sumaCubos 0       = 0\"\r\n| \"sumaCubos (Suc n) = sumaCubos n + (Suc n)^3\"\r\n\r\n(* 1\u00aa demostraci\u00f3n *)\r\nlemma\r\n  \"4 * sumaCubos n = (n*(n+1))^2\"\r\nproof (induct n)\r\n  show \"4 * sumaCubos 0 = (0 * (0 + 1))^2\"\r\n    by simp\r\nnext\r\n  fix n\r\n  assume HI : \"4 * sumaCubos n = (n * (n + 1))^2\"\r\n  have \"4 * sumaCubos (Suc n) = 4 * (sumaCubos n + (n+1)^3)\"\r\n    by simp\r\n  also have \"\u2026 = 4 * sumaCubos n + 4*(n+1)^3\"\r\n    by simp\r\n  also have \"\u2026 = (n*(n+1))^2 + 4*(n+1)^3\"\r\n    using HI by simp\r\n  also have \"\u2026 = (n+1)^2*(n^2+4*n+4)\"\r\n    by algebra\r\n  also have \"\u2026 = (n+1)^2*(n+2)^2\"\r\n    by algebra\r\n  also have \"\u2026 = ((n+1)*((n+1)+1))^2\"\r\n    by algebra\r\n  also have \"\u2026 = (Suc n * (Suc n + 1))^2\"\r\n    by (simp only: Suc_eq_plus1)\r\n  finally show \"4 * sumaCubos (Suc n) = (Suc n * (Suc n + 1))^2\"\r\n    by this\r\nqed\r\n\r\n(* 2\u00aa demostraci\u00f3n *)\r\nlemma\r\n  \"4 * sumaCubos n = (n*(n+1))^2\"\r\nproof (induct n)\r\n  show \"4 * sumaCubos 0 = (0 * (0 + 1))^2\"\r\n    by simp\r\nnext\r\n  fix n\r\n  assume HI : \"4 * sumaCubos n = (n * (n + 1))^2\"\r\n  have \"4 * sumaCubos (Suc n) = 4 * sumaCubos n + 4*(n+1)^3\"\r\n    by simp\r\n  also have \"\u2026 = (n*(n+1))^2 + 4*(n+1)^3\"\r\n    using HI by simp\r\n  also have \"\u2026 = ((n+1)*((n+1)+1))^2\"\r\n    by algebra\r\n  also have \"\u2026 = (Suc n * (Suc n + 1))^2\"\r\n    by (simp only: Suc_eq_plus1)\r\n  finally show \"4 * sumaCubos (Suc n) = (Suc n * (Suc n + 1))^2\" .\r\nqed\r\n\r\nend\r\n<\/pre>\n<p>En los comentarios se pueden escribir otras soluciones, escribiendo el c\u00f3digo entre una l\u00ednea con &#60;pre lang=&quot;isar&quot;&#62; y otra con &#60;\/pre&#62;<br \/>\n[\/expand]<\/p>\n","protected":false},"excerpt":{"rendered":"<p>Demostrar que la suma de los primeros cubos 0\u00b3 + 1\u00b3 + 2\u00b3 + 3\u00b3 + \u00b7\u00b7\u00b7 + n\u00b3 es (n(n+1)\/2)\u00b2 Para ello, completar la siguiente teor\u00eda de Lean: import data.nat.basic import tactic open nat variable (n : \u2115) def sumaCubos : \u2115 \u2192 \u2115 | 0 := 0 | (n+1) := sumaCubos n + (n+1)^3 example : 4 * sumaCubos n = (n*(n+1))^2 := sorry [expand title=\u00bbSoluciones con Lean\u00bb] import data.nat.basic import tactic open nat variable (n : \u2115) set_option pp.structure_projections false @[simp] def sumaCubos : \u2115 \u2192 \u2115 | 0 := 0 | (n+1) := sumaCubos n + (n+1)^3 &#8212; 1\u00aa demostraci\u00f3n example : 4 * sumaCubos n = (n*(n+1))^2 := begin induction n with n HI, { simp, ring, }, { calc&#8230;<\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"jetpack_post_was_ever_published":false,"_kad_post_transparent":"","_kad_post_title":"","_kad_post_layout":"","_kad_post_sidebar_id":"","_kad_post_content_style":"","_kad_post_vertical_padding":"","_kad_post_feature":"","_kad_post_feature_position":"","_kad_post_header":false,"_kad_post_footer":false,"_jetpack_newsletter_access":"","_jetpack_dont_email_post_to_subs":false,"_jetpack_newsletter_tier_id":0,"_jetpack_memberships_contains_paywalled_content":false,"_jetpack_memberships_contains_paid_content":false,"footnotes":""},"categories":[280],"tags":[],"jetpack_featured_media_url":"","jetpack_sharing_enabled":true,"_links":{"self":[{"href":"https:\/\/www.glc.us.es\/~jalonso\/calculemus\/wp-json\/wp\/v2\/posts\/779"}],"collection":[{"href":"https:\/\/www.glc.us.es\/~jalonso\/calculemus\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/www.glc.us.es\/~jalonso\/calculemus\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/www.glc.us.es\/~jalonso\/calculemus\/wp-json\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/www.glc.us.es\/~jalonso\/calculemus\/wp-json\/wp\/v2\/comments?post=779"}],"version-history":[{"count":1,"href":"https:\/\/www.glc.us.es\/~jalonso\/calculemus\/wp-json\/wp\/v2\/posts\/779\/revisions"}],"predecessor-version":[{"id":780,"href":"https:\/\/www.glc.us.es\/~jalonso\/calculemus\/wp-json\/wp\/v2\/posts\/779\/revisions\/780"}],"wp:attachment":[{"href":"https:\/\/www.glc.us.es\/~jalonso\/calculemus\/wp-json\/wp\/v2\/media?parent=779"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.glc.us.es\/~jalonso\/calculemus\/wp-json\/wp\/v2\/categories?post=779"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.glc.us.es\/~jalonso\/calculemus\/wp-json\/wp\/v2\/tags?post=779"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}