{"id":777,"date":"2021-09-21T05:00:31","date_gmt":"2021-09-21T03:00:31","guid":{"rendered":"https:\/\/www.glc.us.es\/~jalonso\/calculemus\/?p=777"},"modified":"2021-09-16T16:28:04","modified_gmt":"2021-09-16T14:28:04","slug":"suma-de-los-primeros-cuadrados","status":"publish","type":"post","link":"https:\/\/www.glc.us.es\/~jalonso\/calculemus\/suma-de-los-primeros-cuadrados\/","title":{"rendered":"Suma de los primeros cuadrados"},"content":{"rendered":"

Demostrar que la suma de los primeros cuadrados<\/p>\n

\n   0\u00b2 + 1\u00b2 + 2\u00b2 + 3\u00b2 + \u00b7\u00b7\u00b7 + n\u00b2\n<\/pre>\n
es n(n+1)(2n+1)\/6.<\/p>\n
Para ello, completar la siguiente teor\u00eda de Lean:<\/p>\n
\nimport data.nat.basic\nimport tactic\nopen nat\n\nvariable (n : \u2115)\n\n@[simp]\ndef sumaCuadrados : \u2115 \u2192 \u2115\n| 0     := 0\n| (n+1) := sumaCuadrados n + (n+1)^2\n\nexample :\n  6 * sumaCuadrados n = n* (n + 1) * (2 * n + 1) :=\nsorry\n<\/pre>\n
[expand title=\u00bbSoluciones con Lean\u00bb]<\/p>\n
\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 sumaCuadrados : \u2115 \u2192 \u2115\r\n| 0     := 0\r\n| (n+1) := sumaCuadrados n + (n+1)^2\r\n\r\n-- 1\u00aa demostraci\u00f3n\r\nexample :\r\n  6 * sumaCuadrados n = n* (n + 1) * (2 * n + 1) :=\r\nbegin\r\n  induction n with n HI,\r\n  { simp, },\r\n  { calc 6 * sumaCuadrados (succ n)\r\n         = 6 * (sumaCuadrados n + (n+1)^2)\r\n           : by simp\r\n     ... = 6 * sumaCuadrados n + 6 * (n+1)^2\r\n           : by ring_nf\r\n     ... = n * (n + 1) * (2 * n + 1) + 6 * (n+1)^2\r\n           : by {congr; rw HI}\r\n     ... = (n + 1) * (n * (2 * n + 1) + 6 * (n+1))\r\n           : by ring_nf\r\n     ... = (n + 1) * (2 * n^2 + 7 * n + 6)\r\n           : by ring_nf\r\n     ... = (n + 1) * (n + 2) * (2 * n + 3)\r\n           : by ring_nf\r\n     ... = succ n * (succ n + 1) * (2 * succ n + 1)\r\n           : by refl, },\r\nend\r\n\r\n-- 2\u00aa demostraci\u00f3n\r\nexample :\r\n  6 * sumaCuadrados n = n* (n + 1) * (2 * n + 1) :=\r\nbegin\r\n  induction n with n HI,\r\n  { simp, },\r\n  { calc 6 * sumaCuadrados (succ n)\r\n         = 6 * (sumaCuadrados n + (n+1)^2)\r\n           : by simp\r\n     ... = 6 * sumaCuadrados n + 6 * (n+1)^2\r\n           : by ring_nf\r\n     ... = n * (n + 1) * (2 * n + 1) + 6 * (n+1)^2\r\n           : by {congr; rw HI}\r\n     ... = (n + 1) * (n + 2) * (2 * n + 3)\r\n           : by ring_nf\r\n     ... = succ n * (succ n + 1) * (2 * succ n + 1)\r\n           : by refl, },\r\nend\r\n<\/pre>\n
Se puede interactuar con la prueba anterior en esta sesi\u00f3n con Lean<\/a>.<\/p>\n
En los comentarios se pueden escribir otras soluciones, escribiendo el c\u00f3digo entre una l\u00ednea con <pre lang="lean"> y otra con <\/pre>
\n[\/expand]<\/p>\n
[expand title=\u00bbSoluciones con Isabelle\/HOL\u00bb]<\/p>\n
\r\ntheory Suma_de_los_primeros_cuadrados\r\nimports Main\r\nbegin\r\n\r\nfun sumaCuadrados :: \"nat \u21d2 nat\" where\r\n  \"sumaCuadrados 0       = 0\"\r\n| \"sumaCuadrados (Suc n) = sumaCuadrados n + (Suc n)^2\"\r\n\r\n(* 1\u00aa demostraci\u00f3n *)\r\nlemma\r\n  \"6 * sumaCuadrados n = n* (n + 1) * (2 * n + 1)\"\r\nproof (induct n)\r\n  show \"6 * sumaCuadrados 0 =  0 * (0 + 1) * (2 * 0 + 1)\"\r\n    by simp\r\nnext\r\n  fix n\r\n  assume HI : \"6 * sumaCuadrados n = n * (n + 1) * (2 * n + 1)\"\r\n  have \"6 * sumaCuadrados (Suc n) =\r\n        6 * (sumaCuadrados n + (n + 1)^2)\"\r\n    by simp\r\n  also have \"\u2026 = 6 * sumaCuadrados n + 6 * (n + 1)^2\"\r\n    by simp\r\n  also have \"\u2026 = n * (n + 1) * (2 * n + 1) + 6 * (n + 1)^2\"\r\n    using HI by simp\r\n  also have \"\u2026 = (n + 1) * (n * (2 * n + 1) + 6 * (n+1))\"\r\n    by algebra\r\n  also have \"\u2026 = (n + 1) * (2 * n^2 + 7 * n + 6)\"\r\n    by algebra\r\n  also have \"\u2026 = (n + 1) * (n + 2) * (2 * n + 3)\"\r\n    by algebra\r\n  also have \"\u2026 = (n + 1) * ((n + 1) + 1) * (2 * (n + 1) + 1)\"\r\n    by algebra\r\n  also have \"\u2026 = Suc n * (Suc n + 1) * (2 * Suc n + 1)\"\r\n    by (simp only: Suc_eq_plus1)\r\n  finally show \"6 * sumaCuadrados (Suc n) =\r\n        Suc n * (Suc n + 1) * (2 * Suc n + 1)\"\r\n    by this\r\nqed\r\n\r\n(* 2\u00aa demostraci\u00f3n *)\r\nlemma\r\n  \"6 * sumaCuadrados n = n* (n + 1) * (2 * n + 1)\"\r\nproof (induct n)\r\n  show \"6 * sumaCuadrados 0 =  0 * (0 + 1) * (2 * 0 + 1)\"\r\n    by simp\r\nnext\r\n  fix n\r\n  assume HI : \"6 * sumaCuadrados n = n * (n + 1) * (2 * n + 1)\"\r\n  have \"6 * sumaCuadrados (Suc n) =\r\n        6 * sumaCuadrados n + 6 * (n + 1)^2\"\r\n    by simp\r\n  also have \"\u2026 = n * (n + 1) * (2 * n + 1) + 6 * (n + 1)^2\"\r\n    using HI by simp\r\n  also have \"\u2026 = (n + 1) * ((n + 1) + 1) * (2 * (n + 1) + 1)\"\r\n    by algebra\r\n  finally show \"6 * sumaCuadrados (Suc n) =\r\n        Suc n * (Suc n + 1) * (2 * Suc n + 1)\"\r\n    by (simp only: Suc_eq_plus1)\r\nqed\r\n\r\nend\r\n<\/pre>\n
En los comentarios se pueden escribir otras soluciones, escribiendo el c\u00f3digo entre una l\u00ednea con <pre lang="isar"> y otra con <\/pre>
\n[\/expand]<\/p>\n","protected":false},"excerpt":{"rendered":"
Demostrar que la suma de los primeros cuadrados 0\u00b2 + 1\u00b2 + 2\u00b2 + 3\u00b2 + \u00b7\u00b7\u00b7 + n\u00b2 es n(n+1)(2n+1)\/6. Para ello, completar la siguiente teor\u00eda de Lean: import data.nat.basic import tactic open nat variable (n : \u2115) @[simp] def sumaCuadrados : \u2115 \u2192 \u2115 | 0 := 0 | (n+1) := sumaCuadrados n + (n+1)^2 example : 6 * sumaCuadrados n = n* (n + 1) * (2 * n + 1) := 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 sumaCuadrados : \u2115 \u2192 \u2115 | 0 := 0 | (n+1) := sumaCuadrados n + (n+1)^2 — 1\u00aa demostraci\u00f3n example : 6 * sumaCuadrados n = n* (n + 1)…<\/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\/777"}],"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=777"}],"version-history":[{"count":1,"href":"https:\/\/www.glc.us.es\/~jalonso\/calculemus\/wp-json\/wp\/v2\/posts\/777\/revisions"}],"predecessor-version":[{"id":778,"href":"https:\/\/www.glc.us.es\/~jalonso\/calculemus\/wp-json\/wp\/v2\/posts\/777\/revisions\/778"}],"wp:attachment":[{"href":"https:\/\/www.glc.us.es\/~jalonso\/calculemus\/wp-json\/wp\/v2\/media?parent=777"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.glc.us.es\/~jalonso\/calculemus\/wp-json\/wp\/v2\/categories?post=777"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.glc.us.es\/~jalonso\/calculemus\/wp-json\/wp\/v2\/tags?post=777"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}