{"id":34747,"date":"2019-08-24T16:52:05","date_gmt":"2019-08-24T15:52:05","guid":{"rendered":"https:\/\/groupe-reussite.fr\/?page_id=34747"},"modified":"2022-07-12T12:07:03","modified_gmt":"2022-07-12T11:07:03","slug":"cours-en-ligne-corriges-sommes-produits-maths-sup","status":"publish","type":"page","link":"https:\/\/groupe-reussite.fr\/ressources\/cours-en-ligne-corriges-sommes-produits-maths-sup\/","title":{"rendered":"Sommes et produits – Corrig\u00e9s"},"content":{"rendered":"
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\n\t\t\t\t\t\t\t\t<\/div>\n\t\t\t\t\t<\/div>\n\t\t<\/div>\n\t\t\t\t\t\t\t\t<\/div>\n\t\t\t\t\t<\/div>\n\t\t<\/section>\n\t\t\t\t
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Cours en ligne Maths en Maths Sup<\/a><\/p>\t\t<\/div>\n\t\t\t\t<\/div>\n\t\t\t\t

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Chapitres Maths en MPSI, PCSI, MP2I, PTSI<\/p>\t\t<\/div>\n\t\t\t\t<\/div>\n\t\t\t\t

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\n\t\t\t\n\t\t\t\t\t\t\n\t\t\t\t\t\tR\u00e9currence<\/span>\n\t\t<\/span>\n\t\t\t\t\t<\/a>\n\t\t<\/div>\n\t\t\t\t<\/div>\n\t\t\t\t<\/div>\n\t\t\t\t
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\n\t\t\t\n\t\t\t\t\t\t\n\t\t\t\t\t\tSommes et produits<\/span>\n\t\t<\/span>\n\t\t\t\t\t<\/a>\n\t\t<\/div>\n\t\t\t\t<\/div>\n\t\t\t\t<\/div>\n\t\t\t\t
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\n\t\t\t\n\t\t\t\t\t\t\n\t\t\t\t\t\tNombres complexes<\/span>\n\t\t<\/span>\n\t\t\t\t\t<\/a>\n\t\t<\/div>\n\t\t\t\t<\/div>\n\t\t\t\t<\/div>\n\t\t\t\t
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\n\t\t\t\n\t\t\t\t\t\t\n\t\t\t\t\t\tTrigonom\u00e9trie<\/span>\n\t\t<\/span>\n\t\t\t\t\t<\/a>\n\t\t<\/div>\n\t\t\t\t<\/div>\n\t\t\t\t<\/div>\n\t\t\t\t
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\n\t\t\t\n\t\t\t\t\t\t\n\t\t\t\t\t\tNombres r\u00e9els<\/span>\n\t\t<\/span>\n\t\t\t\t\t<\/a>\n\t\t<\/div>\n\t\t\t\t<\/div>\n\t\t\t\t<\/div>\n\t\t\t\t
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\n\t\t\t\n\t\t\t\t\t\t\n\t\t\t\t\t\tEnsembles et applications<\/span>\n\t\t<\/span>\n\t\t\t\t\t<\/a>\n\t\t<\/div>\n\t\t\t\t<\/div>\n\t\t\t\t<\/div>\n\t\t\t\t
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\n\t\t\t\n\t\t\t\t\t\t\n\t\t\t\t\t\tFonctions (intro)<\/span>\n\t\t<\/span>\n\t\t\t\t\t<\/a>\n\t\t<\/div>\n\t\t\t\t<\/div>\n\t\t\t\t<\/div>\n\t\t\t\t
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\n\t\t\t\n\t\t\t\t\t\t\n\t\t\t\t\t\tFonctions usuelles<\/span>\n\t\t<\/span>\n\t\t\t\t\t<\/a>\n\t\t<\/div>\n\t\t\t\t<\/div>\n\t\t\t\t<\/div>\n\t\t\t\t
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\n\t\t\t\n\t\t\t\t\t\t\n\t\t\t\t\t\tPrimitives<\/span>\n\t\t<\/span>\n\t\t\t\t\t<\/a>\n\t\t<\/div>\n\t\t\t\t<\/div>\n\t\t\t\t<\/div>\n\t\t\t\t
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\n\t\t\t\n\t\t\t\t\t\t\n\t\t\t\t\t\t\u00c9quations diff\u00e9rentielles<\/span>\n\t\t<\/span>\n\t\t\t\t\t<\/a>\n\t\t<\/div>\n\t\t\t\t<\/div>\n\t\t\t\t<\/div>\n\t\t\t\t
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\n\t\t\t\n\t\t\t\t\t\t\n\t\t\t\t\t\tSuites num\u00e9riques<\/span>\n\t\t<\/span>\n\t\t\t\t\t<\/a>\n\t\t<\/div>\n\t\t\t\t<\/div>\n\t\t\t\t<\/div>\n\t\t\t\t
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\n\t\t\t\n\t\t\t\t\t\t\n\t\t\t\t\t\tLimites et continuit\u00e9<\/span>\n\t\t<\/span>\n\t\t\t\t\t<\/a>\n\t\t<\/div>\n\t\t\t\t<\/div>\n\t\t\t\t<\/div>\n\t\t\t\t
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\n\t\t\t\n\t\t\t\t\t\t\n\t\t\t\t\t\tD\u00e9riv\u00e9es<\/span>\n\t\t<\/span>\n\t\t\t\t\t<\/a>\n\t\t<\/div>\n\t\t\t\t<\/div>\n\t\t\t\t<\/div>\n\t\t\t\t
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\n\t\t\t\n\t\t\t\t\t\t\n\t\t\t\t\t\tSyst\u00e8mes<\/span>\n\t\t<\/span>\n\t\t\t\t\t<\/a>\n\t\t<\/div>\n\t\t\t\t<\/div>\n\t\t\t\t<\/div>\n\t\t\t\t
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\n\t\t\t\n\t\t\t\t\t\t\n\t\t\t\t\t\tPolyn\u00f4mes<\/span>\n\t\t<\/span>\n\t\t\t\t\t<\/a>\n\t\t<\/div>\n\t\t\t\t<\/div>\n\t\t\t\t<\/div>\n\t\t\t\t
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\n\t\t\t\n\t\t\t\t\t\t\n\t\t\t\t\t\tFractions rationnelles<\/span>\n\t\t<\/span>\n\t\t\t\t\t<\/a>\n\t\t<\/div>\n\t\t\t\t<\/div>\n\t\t\t\t<\/div>\n\t\t\t\t
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\n\t\t\t\n\t\t\t\t\t\t\n\t\t\t\t\t\tArithm\u00e9tique et polyn\u00f4mes<\/span>\n\t\t<\/span>\n\t\t\t\t\t<\/a>\n\t\t<\/div>\n\t\t\t\t<\/div>\n\t\t\t\t<\/div>\n\t\t\t\t
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\n\t\t\t\n\t\t\t\t\t\t\n\t\t\t\t\t\tArithm\u00e9tique et fractions rationnelles<\/span>\n\t\t<\/span>\n\t\t\t\t\t<\/a>\n\t\t<\/div>\n\t\t\t\t<\/div>\n\t\t\t\t<\/div>\n\t\t\t\t
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\n\t\t\t\n\t\t\t\t\t\t\n\t\t\t\t\t\tAnalyse asymptotique<\/span>\n\t\t<\/span>\n\t\t\t\t\t<\/a>\n\t\t<\/div>\n\t\t\t\t<\/div>\n\t\t\t\t<\/div>\n\t\t\t\t
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\n\t\t\t\n\t\t\t\t\t\t\n\t\t\t\t\t\tD\u00e9veloppements limit\u00e9s<\/span>\n\t\t<\/span>\n\t\t\t\t\t<\/a>\n\t\t<\/div>\n\t\t\t\t<\/div>\n\t\t\t\t<\/div>\n\t\t\t\t
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\n\t\t\t\n\t\t\t\t\t\t\n\t\t\t\t\t\tD\u00e9nombrements<\/span>\n\t\t<\/span>\n\t\t\t\t\t<\/a>\n\t\t<\/div>\n\t\t\t\t<\/div>\n\t\t\t\t<\/div>\n\t\t\t\t
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\n\t\t\t\n\t\t\t\t\t\t\n\t\t\t\t\t\tEspaces vectoriels<\/span>\n\t\t<\/span>\n\t\t\t\t\t<\/a>\n\t\t<\/div>\n\t\t\t\t<\/div>\n\t\t\t\t<\/div>\n\t\t\t\t
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\n\t\t\t\n\t\t\t\t\t\t\n\t\t\t\t\t\tMatrices<\/span>\n\t\t<\/span>\n\t\t\t\t\t<\/a>\n\t\t<\/div>\n\t\t\t\t<\/div>\n\t\t\t\t<\/div>\n\t\t\t\t
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\n\t\t\t\n\t\t\t\t\t\t\n\t\t\t\t\t\tInt\u00e9grations<\/span>\n\t\t<\/span>\n\t\t\t\t\t<\/a>\n\t\t<\/div>\n\t\t\t\t<\/div>\n\t\t\t\t<\/div>\n\t\t\t\t
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\n\t\t\t\n\t\t\t\t\t\t\n\t\t\t\t\t\tEspaces pr\u00e9hilbertiens<\/span>\n\t\t<\/span>\n\t\t\t\t\t<\/a>\n\t\t<\/div>\n\t\t\t\t<\/div>\n\t\t\t\t<\/div>\n\t\t\t\t
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\n\t\t\t\n\t\t\t\t\t\t\n\t\t\t\t\t\tEspaces euclidiens<\/span>\n\t\t<\/span>\n\t\t\t\t\t<\/a>\n\t\t<\/div>\n\t\t\t\t<\/div>\n\t\t\t\t<\/div>\n\t\t\t\t
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\n\t\t\t\n\t\t\t\t\t\t\n\t\t\t\t\t\tS\u00e9ries num\u00e9riques<\/span>\n\t\t<\/span>\n\t\t\t\t\t<\/a>\n\t\t<\/div>\n\t\t\t\t<\/div>\n\t\t\t\t<\/div>\n\t\t\t\t
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\n\t\t\t\n\t\t\t\t\t\t\n\t\t\t\t\t\tProbabilit\u00e9s<\/span>\n\t\t<\/span>\n\t\t\t\t\t<\/a>\n\t\t<\/div>\n\t\t\t\t<\/div>\n\t\t\t\t<\/div>\n\t\t\t\t
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\n\t\t\t\n\t\t\t\t\t\t\n\t\t\t\t\t\tD\u00e9terminants<\/span>\n\t\t<\/span>\n\t\t\t\t\t<\/a>\n\t\t<\/div>\n\t\t\t\t<\/div>\n\t\t\t\t<\/div>\n\t\t\t\t
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\n\t\t\t\n\t\t\t\t\t\t\n\t\t\t\t\t\tVariables al\u00e9atoires<\/span>\n\t\t<\/span>\n\t\t\t\t\t<\/a>\n\t\t<\/div>\n\t\t\t\t<\/div>\n\t\t\t\t<\/div>\n\t\t\t\t
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\n\t\t\t\n\t\t\t\t\t\t\n\t\t\t\t\t\tCONTACTEZ-NOUS<\/span>\n\t\t<\/span>\n\t\t\t\t\t<\/a>\n\t\t<\/div>\n\t\t\t\t<\/div>\n\t\t\t\t<\/div>\n\t\t\t\t\t\t<\/div>\n\t\t\t\t\t<\/div>\n\t\t<\/div>\n\t\t\t\t\t\t\t\t<\/div>\n\t\t\t\t\t<\/div>\n\t\t<\/section>\n\t\t\t\t
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Cours en ligne Maths en Maths Sup<\/a><\/p>\t\t<\/div>\n\t\t\t\t<\/div>\n\t\t\t\t

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Chapitres Maths en MPSI, PCSI, MP2I, PTSI<\/p>\t\t<\/div>\n\t\t\t\t<\/div>\n\t\t\t\t

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\n\t\t\t\n\t\t\t\t\t\t\n\t\t\t\t\t\tR\u00e9currence<\/span>\n\t\t<\/span>\n\t\t\t\t\t<\/a>\n\t\t<\/div>\n\t\t\t\t<\/div>\n\t\t\t\t<\/div>\n\t\t\t\t
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\n\t\t\t\n\t\t\t\t\t\t\n\t\t\t\t\t\tSommes et produits<\/span>\n\t\t<\/span>\n\t\t\t\t\t<\/a>\n\t\t<\/div>\n\t\t\t\t<\/div>\n\t\t\t\t<\/div>\n\t\t\t\t
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\n\t\t\t\n\t\t\t\t\t\t\n\t\t\t\t\t\tNombres complexes<\/span>\n\t\t<\/span>\n\t\t\t\t\t<\/a>\n\t\t<\/div>\n\t\t\t\t<\/div>\n\t\t\t\t<\/div>\n\t\t\t\t
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\n\t\t\t\n\t\t\t\t\t\t\n\t\t\t\t\t\tTrigonom\u00e9trie<\/span>\n\t\t<\/span>\n\t\t\t\t\t<\/a>\n\t\t<\/div>\n\t\t\t\t<\/div>\n\t\t\t\t<\/div>\n\t\t\t\t
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\n\t\t\t\n\t\t\t\t\t\t\n\t\t\t\t\t\tNombres r\u00e9els<\/span>\n\t\t<\/span>\n\t\t\t\t\t<\/a>\n\t\t<\/div>\n\t\t\t\t<\/div>\n\t\t\t\t<\/div>\n\t\t\t\t
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\n\t\t\t\n\t\t\t\t\t\t\n\t\t\t\t\t\tEnsembles et applications<\/span>\n\t\t<\/span>\n\t\t\t\t\t<\/a>\n\t\t<\/div>\n\t\t\t\t<\/div>\n\t\t\t\t<\/div>\n\t\t\t\t
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\n\t\t\t\n\t\t\t\t\t\t\n\t\t\t\t\t\tFonctions (intro)<\/span>\n\t\t<\/span>\n\t\t\t\t\t<\/a>\n\t\t<\/div>\n\t\t\t\t<\/div>\n\t\t\t\t<\/div>\n\t\t\t\t
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\n\t\t\t\n\t\t\t\t\t\t\n\t\t\t\t\t\tFonctions usuelles<\/span>\n\t\t<\/span>\n\t\t\t\t\t<\/a>\n\t\t<\/div>\n\t\t\t\t<\/div>\n\t\t\t\t<\/div>\n\t\t\t\t
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\n\t\t\t\n\t\t\t\t\t\t\n\t\t\t\t\t\tPrimitives<\/span>\n\t\t<\/span>\n\t\t\t\t\t<\/a>\n\t\t<\/div>\n\t\t\t\t<\/div>\n\t\t\t\t<\/div>\n\t\t\t\t
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\n\t\t\t\n\t\t\t\t\t\t\n\t\t\t\t\t\t\u00c9quations diff\u00e9rentielles<\/span>\n\t\t<\/span>\n\t\t\t\t\t<\/a>\n\t\t<\/div>\n\t\t\t\t<\/div>\n\t\t\t\t<\/div>\n\t\t\t\t
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\n\t\t\t\n\t\t\t\t\t\t\n\t\t\t\t\t\tSuites num\u00e9riques<\/span>\n\t\t<\/span>\n\t\t\t\t\t<\/a>\n\t\t<\/div>\n\t\t\t\t<\/div>\n\t\t\t\t<\/div>\n\t\t\t\t
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\n\t\t\t\n\t\t\t\t\t\t\n\t\t\t\t\t\tLimites et continuit\u00e9<\/span>\n\t\t<\/span>\n\t\t\t\t\t<\/a>\n\t\t<\/div>\n\t\t\t\t<\/div>\n\t\t\t\t<\/div>\n\t\t\t\t
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\n\t\t\t\n\t\t\t\t\t\t\n\t\t\t\t\t\tD\u00e9riv\u00e9es<\/span>\n\t\t<\/span>\n\t\t\t\t\t<\/a>\n\t\t<\/div>\n\t\t\t\t<\/div>\n\t\t\t\t<\/div>\n\t\t\t\t
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\n\t\t\t\n\t\t\t\t\t\t\n\t\t\t\t\t\tSyst\u00e8mes<\/span>\n\t\t<\/span>\n\t\t\t\t\t<\/a>\n\t\t<\/div>\n\t\t\t\t<\/div>\n\t\t\t\t<\/div>\n\t\t\t\t
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\n\t\t\t\n\t\t\t\t\t\t\n\t\t\t\t\t\tPolyn\u00f4mes<\/span>\n\t\t<\/span>\n\t\t\t\t\t<\/a>\n\t\t<\/div>\n\t\t\t\t<\/div>\n\t\t\t\t<\/div>\n\t\t\t\t
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\n\t\t\t\n\t\t\t\t\t\t\n\t\t\t\t\t\tFractions rationnelles<\/span>\n\t\t<\/span>\n\t\t\t\t\t<\/a>\n\t\t<\/div>\n\t\t\t\t<\/div>\n\t\t\t\t<\/div>\n\t\t\t\t
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\n\t\t\t\n\t\t\t\t\t\t\n\t\t\t\t\t\tArithm\u00e9tique et polyn\u00f4mes<\/span>\n\t\t<\/span>\n\t\t\t\t\t<\/a>\n\t\t<\/div>\n\t\t\t\t<\/div>\n\t\t\t\t<\/div>\n\t\t\t\t
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\n\t\t\t\n\t\t\t\t\t\t\n\t\t\t\t\t\tArithm\u00e9tique et fractions rationnelles<\/span>\n\t\t<\/span>\n\t\t\t\t\t<\/a>\n\t\t<\/div>\n\t\t\t\t<\/div>\n\t\t\t\t<\/div>\n\t\t\t\t
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Corrig\u00e9s : Sommes et produits<\/h1>\n<\/div>\n

R\u00e9sum\u00e9 de cours<\/a> Exercices<\/a> Corrig\u00e9s<\/a><\/p>\n

Cours en ligne de Maths en Maths Sup<\/a><\/p>\n

Corrig\u00e9s – sommes et produits MPSI, PCSI<\/h2>\n

1. QCM – Vrai\/Faux<\/h3>\n

Les relations suivantes sont- elles vraies ? Les corriger lorsqu’elles sont fausses.<\/p>\n

Question 1<\/b>
\nSi \"n\in \mathbb{N}\", \"\displaystyle \sum _{k = 0} ^n \binom n k (- 1) ^k = 0\".<\/p>\n

Corrig\u00e9 :\u00a0<\/strong><\/p>\n

L’affirmation est vraie si \"n > 0\" et fausse pour \"n = 0\".
\nPar le bin\u00f4me de Newton,
\n\"\displaystyle \sum _{k = 0} ^n \binom n k(- 1) ^k \, 1 ^{n - k} = (1 -1) ^n\".
\nLe r\u00e9sultat est nul si \"n \geq 1\" et \u00e9gal \u00e0 1 si \"n = 0\".<\/p>\n

Question 2<\/b>
\nSi \"n\in \mathbb{N}\", \"\displaystyle \sum _{k = 0} ^n \binom n k = 2 ^n\".<\/p>\n

Corrig\u00e9<\/strong> :<\/p>\n

Vrai. Par le bin\u00f4me de Newton,
\n\"\displaystyle \sum _{k = 0} ^n \binom n k 1^k\, 1 ^{n - k} = (1 +1) ^n = 2 ^n\".<\/p>\n

Question 3<\/b>
\nSoit \"n \in \mathbb{N}^*\". \"\displaystyle \sum _{k = 0} ^{\lfloor n /2 \rfloor } \binom n {2 k + 1} = 2 ^{n - 1}\".<\/p>\n

Corrig\u00e9<\/strong> :<\/p>\n

Faux. La r\u00e9ponse correcte est
\n\"B =\displaystyle \sum _{k = 0} ^{\lfloor (n - 1) /2 \rfloor } \binom n {2 k + 1} = 2 ^{n-1}\".<\/p>\n

Par le bin\u00f4me de Newton,
\n\"\displaystyle (1 + 1) ^n = \sum _ {p = 0} ^n \binom {n } {p}\"
\n\"\displaystyle (1 - 1 ) ^n = \sum _ {p = 0} ^n \binom {n } {p} \, (-1) ^p\" .
\nPar diff\u00e9rence,
\n\"2 ^n \displaystyle = \sum _ {p = 0} ^n \binom {n } {p} (1 - (- 1)^p)\"
\nIl ne reste que les termes pour \"p = 2\, k +1\" avec \"\displaystyle 0 \leq k \leq {\left \lfloor \frac{n - 1} 2 \right \rfloor}\", donc
\n\"\; \; \displaystyle \sum _ {p = 0} ^{\lfloor (n - 1) /2 \rfloor} \binom {n } {2\,p + 1 } = \frac {2 ^n } 2 = 2 ^{n - 1}\".<\/p>\n

\"\$\"\\$<\/p>\n

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COURS DE MATHS<\/h3>\t\t<\/div>\n\t\t\t\t<\/div>\n\t\t\t\t
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Les meilleurs professeurs particuliers<\/h4>\t\t<\/div>\n\t\t\t\t<\/div>\n\t\t\t\t
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Pour progresser et r\u00e9ussir<\/h4>\t\t<\/div>\n\t\t\t\t<\/div>\n\t\t\t\t
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\n\t\t\t\n\t\t\t\t\t\t\n\t\t\t\t\t\tCours particuliers maths<\/span>\n\t\t<\/span>\n\t\t\t\t\t<\/a>\n\t\t<\/div>\n\t\t\t\t<\/div>\n\t\t\t\t<\/div>\n\t\t\t\t
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Avis Google France <\/b>\u2605\u2605\u2605\u2605\u2605<\/b><\/span> 4,9 sur 5<\/b><\/p>\t\t\t\t\t<\/div>\n\t\t\t\t\t\t<\/div>\n\t\t\t\t<\/div>\n\t\t\t\t\t\t<\/div>\n\t\t\t\t\t<\/div>\n\t\t<\/div>\n\t\t\t\t

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\n\t\t\t\t\t\t\t\t<\/div>\n\t\t\t\t\t<\/div>\n\t\t<\/div>\n\t\t\t\t\t\t\t\t<\/div>\n\t\t\t\t\t<\/div>\n\t\t<\/section>\n\t\t\t\t\t\t\t<\/div>\n\t\t\t\t<\/div>\n\t\t\n

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Question 4<\/b>
\nSoit \"n \in \mathbb{N}\". \"\displaystyle \sum _{k = 0} ^{n}2 ^k = 2 ^{n + 1} - 1\".<\/p>\n

Corrig\u00e9 :\u00a0<\/strong><\/p>\n

Vrai. On utilise si \"q \neq 1\", \"\displaystyle \sum _{k = 0} ^{n}q ^k = \frac {1 - q ^{n + 1}} { 1- q}\"<\/p>\n

Question 5<\/b>
\nSi \"n \in \mathbb{N}\" et \"\textrm{j} =\textrm{e} ^{2 \, \textrm{i} \, \pi /3}\,\",\u00a0 \u00a0\"\displaystyle \sum _{k = 0} ^{3\,n + 2 } j ^k = 0\".<\/p>\n

Corrig\u00e9 :\u00a0<\/strong><\/p>\n

Vrai. On utilise si \"q \neq 1\",
\n\"\quad \quad \displaystyle \sum _{k = 0} ^{3 \, n + 2}q ^k = \frac {1 - q ^{3 n + 3}} { 1- q}\"
\net \"\textrm{j} ^{3 n + 3} = \left ( \textrm{j} ^{3 }\right ) ^{n + 1} = 1^{n + 1} = 1 \".<\/p>\n

Question 6<\/b>
\nSoient \"n \in \mathbb{N}^*\" et \"(a_k)_{1 \leq k \leq n}\" une famille de complexes.
\n\"\quad \quad \quad \displaystyle \sum _{k = 1} ^{n } a_k = \sum _{p = 1} ^{n } a_{n + 1 - p} \,\".<\/p>\n

Corrig\u00e9<\/strong> :<\/p>\n

Vrai. En posant \"k = n + 1 - p\" soit \"p = n + 1 - k\", lorsque \"k\" varie de \"1\" \u00e0 \"n\", \"p\" varie de \"n + 1 - 1\" \u00e0 \"n + 1 - n = 1\"
\n\"\quad \quad \displaystyle \sum _{k = 1} ^{n } a_k = \sum _{p = 1} ^{n } a_{n + 1 - p}\, \".<\/p>\n

Question 7<\/b>
\nSoient \"n \in \mathbb{N}^*\" et \"(u_k)_k\" une suite<\/a> r\u00e9elle ou complexe.
\n\"\displaystyle \sum _ {k = 1} ^n \left ( u_k - 2 u_{k + 1} + u_{k + 2} \right ) =\" \"\quad \quad \quad \quad \quad u_1 + u_2 - u_n - u_{n + 2}\,\".<\/p>\n

Corrig\u00e9 :\u00a0<\/strong><\/p>\n

Faux. On pose \"a _k = u_k - u_{k + 1}\,\".
\nOn remarque que \"\quad u_k - 2 u_{k + 1} + u_{k + 2} = a_{k} - a_{k + 1}\"
\npuis on utilise un t\u00e9lescopage :
\n\"\displaystyle S = \sum _ {k = 1} ^n \left ( u_k - 2 u_{k + 1} + u_{k + 2} \right )\"
\n\"\displaystyle S = \sum _ {k = 1} ^n \left ( a_k - a_{k + 1} \right )\"
\n\"\displaystyle S = \sum _ {k = 1} ^n a_k - \sum _ {k = 1} ^n a_{k + 1}\"
\nen posant \"p = k + 1\",
\n\"\displaystyle S = \sum _ {k = 1} ^n a_k - \sum _ {p = 2} ^{n + 1} a_{p}\"
\n\"S = a_1 - a_{n + 1}\" \"S= u_1 - u_2 - u_{n + 1} + u_{n + 2}\,\".<\/p>\n

Question 8<\/b>
\nSoient \"n \in \mathbb{N}^*\" et \"(u_k)_k\" une suite r\u00e9elle ou complexe.
\n\"\quad \quad \displaystyle \sum _ {k =0} ^n u_{2 k} = \sum _ {k = 0} ^{2 \,n} u_p \,\".<\/p>\n

Corrig\u00e9 :\u00a0<\/strong><\/p>\n

Faux (sauf si les termes \"u_{2 k + 1}\" sont tous nuls).
\nOn a ajout\u00e9 dans la somme tous les termes d’indices \"2\, p + 1\" avec \"0 \leq p \leq n - 1\".<\/p>\n

👍 On retiendra qu’il est impossible d’utiliser un changement d’indice \"k = 2 \, p\" ou \"k = 2\, p + 1\".<\/p>\n

Question 9<\/b>
\nSoit \"n \in \mathbb{N}^*\".\u00a0\"\displaystyle \sum _ {k = 1} ^n k ^3 = \left ( \displaystyle \sum _ {k = 1} ^n k \right ) ^2\".<\/p>\n

Corrig\u00e9 :\u00a0<\/strong><\/p>\n

L’affirmation est vraie car
\n\"\displaystyle \sum _ {k = 1} ^n k ^3 = \frac {n ^2(n + 1)^2} 4\"
\net \"\displaystyle \sum _ {k = 1} ^n k = \frac {n \, (n + 1)} 2\".<\/p>\n

Question 10<\/b>
\nSi \"n \in \mathbb{N}^*\" ,\"\quad\displaystyle \sum_{k = 0 }^{2 n + 1} a_k = \sum_{k =0 }^n a_{2 k} + \sum_{k =1 }^n a_{2 k - 1}\"<\/p>\n

Corrig\u00e9 :\u00a0<\/strong><\/p>\n

Faux. \"\displaystyle S = \sum_{k = 0 }^{2 n + 1} a_k\"
\nOn s\u00e9pare les indices \"k = 2 p\" avec \"0 \leq 2 p \leq 2 n\" des indices \"k = 2 q - 1\" avec \"1 \leq 2 q - 1 \leq 2 n + 1\" ssi \"1 \leq q \leq n + 1\"
\n\"\displaystyle S = \sum_{p = 0 }^{n} a_{2 p} + \sum_{q = 1 }^{n + 1}a_{2 q - 1}\"<\/p>\n

Question 11<\/b>
\nSoient \"n \in \mathbb{N}^*\" et \"(a_k)_{1 \leq k \leq n}\" , \"(b_k)_{1 \leq k \leq n}\" deux familles r\u00e9elles ou complexes
\n\"\quad \displaystyle \left (\sum _ {k = 1} ^n a_k \right ) \, \left (\sum _ {k = 1} ^n b_k \right ) =\sum _ {k = 1} ^n a_k \, b_k\,\".<\/p>\n

Corrig\u00e9 :\u00a0<\/strong><\/p>\n

Faux car il manque les termes \"a _ i \, b_j\" si \"i \neq j\"
\n\"\displaystyle \left (\sum _ {k = 1} ^n a_k \right ) \, \left (\sum _ {k = 1} ^n b_k \right ) = \sum _ {i = 1} ^n \sum _ {j = 1} ^n a_i \, b_j\"<\/p>\n

👍 Pour \u00e9viter une telle erreur prendre la pr\u00e9caution d’utiliser des indices diff\u00e9rents dans les deux sommes :
\nc’est \u00e0 dire \u00e9crire
\n\"\displaystyle \left (\sum _ {k = 1} ^n a_k \right ) \, \left (\sum _ {k = 1} ^n b_k \right ) =\" \"\displaystyle \quad \quad\quad \quad \left (\sum _ {i = 1} ^n a_i \right ) \, \left (\sum _ {j = 1} ^n b_j \right )\".<\/p>\n

Question 12<\/b>
\nSoit \"n \in \mathbb{N}^*\",\u00a0\u00a0\"\quad \quad \quad \displaystyle \sum _ {k = 1} ^n \sum _{j = k} ^n 1 = \frac {n \, (n + 1)} 2\"<\/p>\n

Corrig\u00e9 :\u00a0<\/strong><\/p>\n

Vrai. \"\displaystyle S = \sum _ {k = 1} ^n \sum _{j = k} ^n 1\"
\n\"\displaystyle \sum _{j = k} ^n 1 = n - (k - 1) = n - k + 1\"
\n\"\displaystyle S = \sum _ {k = 1} ^n (n - k + 1)\"
\nen posant \"p = n - k + 1\" qui varie de \"1\" \u00e0 \"n\" si \"k\" varie de \"1\" \u00e0 \"n\" :
\n\"\displaystyle S = \sum _ {p = 1} ^n p = \frac {n \, (n + 1)} 2\".<\/p>\n

Question 13<\/b>
\nSi \"n \in \mathbb{N}^*,\displaystyle \sum_{k = - n }^n \frac 1 {k ^2 +1} = 2 \sum_{k =0 }^n \frac 1 {k ^2 +1}\"<\/p>\n

Corrig\u00e9 :\u00a0<\/strong><\/p>\n

\"\displaystyle S_n = \sum_{k = - n }^n \frac 1 {k ^2 +1}\"
\n\"\displaystyle S_n = \sum_{k = - n }^{-1} \frac 1 {k ^2 +1} + \sum_{k = 0 }^n \frac 1 {k ^2 +1}\"
\nen posant \"j = - k\" dans la premi\u00e8re somme
\n\"\displaystyle S_n = \sum_{j = 1 }^{n} \frac 1 {(-j) ^2 +1} + \sum_{k = 0 }^n \frac 1 {k ^2 +1}\"
\n\"\displaystyle S_n = \sum_{j = 1 }^{n} \frac 1 {j ^2 +1} + \sum_{k = 0 }^n \frac 1 {k ^2 +1}\"
\n\"\displaystyle S_n = 2 \, \sum_{k = 0 }^n \frac 1 {k ^2 +1} - 1\".<\/p>\n

Question 14<\/b>
\nSoit \"n \in \mathbb{N}^*\".
\nSi \"x \in \mathbb{R}^+, \, f(x) = \displaystyle \sum _{k = 1} ^n \frac 1 {k + x}\".
\na)\u00a0<\/b>Si \"k \in [\![1, \, n]\!]\", \"\displaystyle f(k) = \sum_{k = 1} ^n \frac 1 {2 \, k}\".<\/p>\n

Corrig\u00e9 :\u00a0<\/strong><\/p>\n

Faux. ⚠️ Si l’expression \"f(x)\" est exprim\u00e9e \u00e0 l’aide d’un signe \"\sum\" ou \"\prod\", pour calculer \"f(k)\" o\u00f9 \"k\" est un entier, \"k\" ne doit pas \u00eatre l’indice de sommation ou de produit. Si n\u00e9cessaire, changer le nom de l’indice de sommation ou de produit.
\nIl faut noter aussi que la relation propos\u00e9e n’a pas de sens : dans \"f(k)\", l’entier \"k\" est fix\u00e9 alors que dans la somme, il est variable<\/a> !<\/p>\n

On \u00e9crit donc \"f(x) = \displaystyle \sum _{j = 1} ^n \frac 1 {j + x}\"
\ndonc \"f(k) = \displaystyle \sum _{j = 1} ^n \frac 1 {j + k}\"
\nen posant \"p = j + k\", \"f(k) = \displaystyle \sum _{p = k+1} ^{n+ k} \frac 1 {p}\".<\/p>\n

b)<\/b>\u00a0\"f(-2 n) + f(n) = \displaystyle \frac 1 {2 n}\".<\/p>\n

Faux. La question pr\u00e9c\u00e9dente donne \"f(n) = \displaystyle \sum _{p = n+1} ^{2n} \frac 1 {p}\".
\n\"f(-2 n) = \displaystyle \sum _{k = 1} ^{n} \frac 1 {k - 2 \,n}\"
\nen posant \"j = 2 \, n - k\", \"j\" varie de \"n\" \u00e0 \"2\, n - 1\" lorsque \"k\" varie de \"1\" \u00e0 \"n\".
\n\"f(-2 \,n) = \displaystyle \sum _{j = n} ^{2\, n - 1 } \frac 1 {-j} = - \sum _{j = n} ^{2\, n - 1 } \frac 1 {j}\"<\/p>\n

\"f(-2 \, n) + f(n) = - \displaystyle \sum _{j = n} ^{2\, n - 1 } \frac 1 {j} + \sum _{p = n+1} ^{2n} \frac 1 {p}\"
\n\"f(-2\, n) + f(n) = - \displaystyle \frac 1 n + \frac {1} {2\, n} = - \frac 1 {2\, n}\" .<\/p>\n

Question 15<\/b>
\nSoient \"n \in \mathbb{N}^*\" et \"(a_{i , j})_{0\leq i , j \leq n}\" une famille r\u00e9elle ou complexe<\/a>.
\n\"\quad \quad \displaystyle \sum _ {i = 0}^n \sum _ {j = i} ^n a_{i ,j} = \sum _ {j = i}^n \sum _ {i = 0} ^n a_{i ,j}\, \".<\/p>\n

Corrig\u00e9 :\u00a0<\/strong><\/p>\n

Faux et n’a d’ailleurs pas de sens, la somme \u00ab\u00a0ext\u00e9rieure \u00a0\u00bb est impossible \u00e0 calculer car \"i\" n’est pas d\u00e9fini.<\/p>\n

\"A = \displaystyle \sum _ {i = 0}^n \sum _ {j = i} ^n a_{i ,j}\"
\nest \u00e9crite sous la forme d’une seule somme : \"A = \displaystyle \sum _ {0 \leq i \leq j \leq n} a_{i,j}\"
\nPuis \"j\" \u00e9tant fix\u00e9 dans \"[\! [0 , n]\!]\", \"i\" doit alors varier de \"0\" \u00e0 \"j\" :
\n\"\quad \quad \quad A = \displaystyle \sum_{j = 0} ^{n} \left ( \sum _ {i = 0}^j a_{i,j} \right )\".<\/p>\n

Par contre, on peut intervertir les sommes dans le cas :
\n\"\quad \quad \displaystyle \sum _ {i = 0}^n \sum _ {j = 0} ^p a_{i ,j} = \sum _ {j = 0}^p \sum _ {i = 0} ^n a_{i ,j}\, \"
\nm\u00eame si \"p \neq n\".<\/p>\n

Question 16<\/b>
\nSoient \"n \in \mathbb{N}^*\" et \"x\" et \"y\" des complexes
\n\"x ^{2 n + 1} + y ^{2 n + 1} =\" \"\quad\quad \quad \quad (x + y) \displaystyle \sum _ {k = 0} ^{2 \, n} (- 1) ^k \, x^{k} y ^{2 n - k}\".<\/p>\n

Corrig\u00e9 :\u00a0<\/strong><\/p>\n

On utilise la formule de Bernoulli en remarquant que
\n\"x ^{2 n + 1} + y ^{2 n + 1} = - (- x) ^{2 n + 1} + y ^{2 n + 1}\" avec \"b = - x\" et \"a = y\".
\nOn obtient :
\n\"x ^{2 n + 1} + y ^{2 n + 1} =\" \"\quad \quad \quad \quad (y + x ) \displaystyle \sum _ {k = 0} ^{2 \, n} ( - x)^{k} y ^{2 n - k}\"<\/p>\n

Question 17<\/b>
\nSoient \"p\" et \"n\" des entiers tels que\u00a0 \"1 \leq p < n\",\u00a0\"\displaystyle \prod _{k = p} ^{n} k = \frac {n!} {p!}\"<\/p>\n

Corrig\u00e9 :\u00a0<\/strong><\/p>\n

on a \"\displaystyle \prod _{k = p} ^{n} k = \frac {n!} {(p - 1)!}\"
\ncar \"(p - 1) ! \displaystyle \prod_{k = p} ^n k = \displaystyle \prod _{k = 1} ^n k = n!\,\".<\/p>\n

Question 18<\/b>
\nSoient \"p\" et \"n\" des entiers tels que \"0 \leq p < n\", \"a \in \mathbb{C}^*\" et\"(b_k)_k\"\u00a0 une famille de complexes.
\n\"\quad \quad \quad \displaystyle \prod _{k = p} ^{n} (a \, b_k) = a \prod _{k = p} ^{n} b_k\,\".<\/p>\n

Corrig\u00e9 :\u00a0<\/strong><\/p>\n

Le \"a\" du produit est en facteur de \"n - (p - 1)\" termes donc<\/div>\n
\"\quad \quad \displaystyle \prod _{k = p} ^{n} (a \, b_k) = a^{n - 1 - p}\prod _{k = p} ^{n} b_k\"<\/div>\n
\u00a0⚠️ C’est une faute courante d’oublier de compter le nombre de fois o\u00f9 le facteur \"a\" est \u00ab\u00a0sorti\u00a0\u00bb du produit.<\/div>\n

 <\/p>\n

2. Des sommes et des coefficients du bin\u00f4me<\/b><\/h3>\n

Exercice 1<\/b>
\nSi \"x \in \mathbb{R}\", calculer \"\displaystyle \sum _ {k = 1} ^n \binom {n} {k} \, k \, x^{k - 1}\"
\net \"\displaystyle \sum _ {k = 0} ^n \binom {n} {k} \, \frac {x^{k + 1}} {k + 1}\".<\/p>\n

Corrig\u00e9 :\u00a0<\/strong><\/p>\n

\"\bullet\" En d\u00e9rivant<\/a> \"f\" telle que \"f(x) = \displaystyle \sum _ {k = 0} ^n \binom {n} {k} \, x^{k} = (1 + x) ^n\",
\non obtient
\n\"f'(x) = \displaystyle \sum _ {k = 1} ^n \binom {n} {k} \, k \, x^{k - 1} = n\, (1 + x)^{n - 1}\"<\/p>\n

\"\bullet\" Puis par int\u00e9gration<\/a>,
\n\"\displaystyle \int_0 ^x f(t)\, \textrm{d} \, t = \sum _ {k = 0} ^n \binom {n} {k} \, \frac {x^{k + 1}} {k + 1}\"
\net \"\displaystyle \int_0 ^x f(t)\, \textrm{d} \, t = \left [ \frac {(1 + t)^{n + 1}} {n + 1} \right ] _ 0 ^x\"
\ndonc \"\displaystyle \sum _ {k = 1} ^n \binom {n} {k} \, \frac {x^{k + 1}} {k + 1} = \frac {(1 + x) ^{n + 1} - 1} {n + 1}\".<\/p>\n

Exercice 2<\/b>
\nAppliquer la formule du triangle de Pascal pour calculer \"\displaystyle \sum _ {k = p} ^n \binom {k} {p}\" lorsque \"1 \leq p \leq n\".<\/p>\n

Corrig\u00e9 :\u00a0<\/strong><\/p>\n

\"\displaystyle S_n = \sum _{k = p} ^ n \binom {k}{p}\".
\nPar la formule du triangle de Pascal (valable si \"k > p\"),
\n\"\displaystyle S_n = \binom {p}{p} + \sum _{k = p+1} ^{n} \left ( \binom {k + 1 }{p+ 1 } - \binom {k}{p+ 1 } \right )\"
\n\"\displaystyle \; = 1 + \sum_{k =p+1}^ {n}\binom {k+ 1 }{p+ 1 } - \sum _{ k =p+1}^{n}\binom {k }{p+ 1 }\"
\navec \"i = k + 1\",
\n\"\displaystyle\;= 1+ \sum_{i =p+2}^{ n +1} \binom {i }{p+ 1 } - \sum _{ k =p+1}^{n}\binom {k }{p+ 1 }\"
\n\"\displaystyle S_n = 1+ \binom {n + 1 }{p+ 1 } -\binom {p + 1 }{p+ 1 } = \binom {n + 1} {p + 1}\"<\/p>\n

Exercice 3<\/b>
\nD\u00e9montrer par r\u00e9currence<\/a> que si \"0\leq p \leq n\", \"\displaystyle \sum _ {k = p} ^n \binom {k} {p} = \binom {n + 1 } {p+ 1}\".<\/p>\n

Corrig\u00e9 :\u00a0<\/strong><\/p>\n

Inconv\u00e9nient : il faut conna\u00eetre la formule si elle n’est pas donn\u00e9e alors que la m\u00e9thode de l’exercice 3 ne n\u00e9cessite pas de conna\u00eetre le r\u00e9sultat.<\/div>\n

On note si \"n \geq p\"
\n\"\quad \quad H_n\" : \"\displaystyle \sum _ {k = p} ^n \binom {k} {p} = \binom {n + 1 } {p+ 1}\".<\/p>\n

\"\ast\" Pour \"n = p\", la propri\u00e9t\u00e9 est \u00e9vidente car elle s’\u00e9crit \"1 = 1\".<\/p>\n

\"\ast\" On suppose qu’elle est vraie au rang \"n \geq p\". Par l’hypoth\u00e8se de r\u00e9currence,
\n\"\displaystyle S_{n + 1} = \sum_{k = p} ^{n + 1} \binom {k}{p}= S_n + \binom {n+1 }{p}\" \"\displaystyle S_{n + 1} = \binom {n +1 }{p+1 }+ \binom {n +1 }{p}\"
\nOn termine par la formule du triangle de Pascal : \"\displaystyle S_{n + 1} =\binom {n +2 }{p+ 1 }\".<\/p>\n

La propri\u00e9t\u00e9 est vraie au rang \"n + 1\". Elle a \u00e9t\u00e9 d\u00e9montr\u00e9e par r\u00e9currence.<\/p>\n

Exercice 4<\/b>
\nUtiliser \"\displaystyle \sum _{k = p} ^ n \binom {k}{p} = \binom {n + 1}{p+ 1}\" pour calculer\"\displaystyle \sum _ {k = 1} ^n k\", \"\displaystyle \sum _ {k = 1} ^n k ^2\" et \"\displaystyle \sum _ {k = 1} ^n k^3\".<\/p>\n

Corrig\u00e9 :\u00a0<\/strong><\/p>\n

\"\bullet\" En prenant \"p = 1\", \"\displaystyle \sum _{k = 1} ^ n k = \binom {n + 1}{2} = \frac {n(n + 1)} 2\".<\/p>\n

\"\bullet\" En prenant \"p = 2\",
\n\"\displaystyle \sum _{k = 2} ^ n \frac {k(k - 1)} 2 = \binom {n + 1}{3}\" \"\displaystyle \sum _{k = 2} ^ n \frac {k(k - 1)} 2 = \frac {n(n + 1)(n - 1)} 6\"
\nOn peut commencer la somme \u00e0 \"k = 1\" car le terme est nul pour \"k = 1\"
\nOn obtient apr\u00e8s multiplication par 2 :
\n\"\displaystyle \sum _{k = 1} ^ n (k ^2 - k) = \frac {n(n + 1)(n - 1)} 3\"
\nsoit \"\displaystyle \sum _{k = 1} ^ n {k ^2 } = \sum _{k = 1} ^ n {k } + \frac {n(n + 1)(n - 1)} 3\"
\n\"\displaystyle \sum _{k = 1} ^ n {k ^2 } = \frac {n (n + 1)} 2 + \frac {n(n + 1)(n - 1)} 3\"
\n\"\displaystyle \sum _{k = 1} ^ n {k ^2 } = \frac {n (n + 1)(2 n + 1)} 6\".<\/p>\n

\"\bullet\" En prenant \"p = 3\",
\n\"\displaystyle \sum _{k = 3} ^ n \frac {k(k - 1)(k - 2) } 6 = \binom {n + 1}{4}\"
\n\"\displaystyle\quad \quad \quad = \frac {n(n + 1)(n - 1)(n - 2) } {24}\"
\nOn commence la somme \u00e0 \"k = 1\" car le terme est nul pour \"k = 1\" et \"k = 2\".
\nOn obtient apr\u00e8s multiplication par 6 :
\n\"\displaystyle \sum _{k = 1} ^ n (k ^3 - 3\, k^2 + 2\, k ) =\" \"\displaystyle \quad \quad \quad \quad \frac {n(n + 1)(n - 1)(n - 2) } 4\"
\n\"\displaystyle \sum _{k = 1} ^ n {k ^3} = 3 \sum _{k = 1} ^ n {k ^2 } - 2 \sum _{k = 1} ^ n {k }\"
\n\"\displaystyle +\, \frac {n(n + 1)(n - 1)(n - 2) } 4\"
\n\"\displaystyle \sum _{k = 1} ^ n {k ^3} = \frac {n(n + 1)(2 n + 1)} 2 - n(n + 1)\"
\n\"\displaystyle + \frac {n(n + 1)(n - 1)(n - 2) } 4\"
\nApr\u00e8s calculs :
\n\"\displaystyle \sum _{k = 1} ^ n {k ^3} = \frac {n^2\,(n + 1)^2 } 4\".<\/p>\n

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COURS PARTICULIERS EN LIGNE<\/h3>\t\t<\/div>\n\t\t\t\t<\/div>\n\t\t\t\t
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Nous avons s\u00e9lectionn\u00e9 pour vous les meilleurs professeurs particuliers.<\/h4>\t\t<\/div>\n\t\t\t\t<\/div>\n\t\t\t\t
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POUR ACC\u00c9L\u00c9RER MA PROGRESSION<\/h4>\t\t<\/div>\n\t\t\t\t<\/div>\n\t\t\t\t
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\n\t\t\t\n\t\t\t\t\t\t\n\t\t\t\t\t\tCours particuliers en ligne<\/span>\n\t\t<\/span>\n\t\t\t\t\t<\/a>\n\t\t<\/div>\n\t\t\t\t<\/div>\n\t\t\t\t<\/div>\n\t\t\t\t
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Avis Google France <\/b>\u2605\u2605\u2605\u2605\u2605<\/b><\/span> 4,9 sur 5<\/b><\/p>\t\t\t\t\t<\/div>\n\t\t\t\t\t\t<\/div>\n\t\t\t\t<\/div>\n\t\t\t\t\t\t<\/div>\n\t\t\t\t\t<\/div>\n\t\t<\/div>\n\t\t\t\t

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\n\t\t\t\t\t\t\t\t<\/div>\n\t\t\t\t\t<\/div>\n\t\t<\/div>\n\t\t\t\t\t\t\t\t<\/div>\n\t\t\t\t\t<\/div>\n\t\t<\/section>\n\t\t\t\t\t\t\t<\/div>\n\t\t\t\t<\/div>\n\t\t\n

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3. Des calculs de sommes<\/h3>\n

Exercice 1<\/b>
\nSi \"n > 1\", \"\displaystyle \sum _ {k = n + 1} ^{2 n} \frac 1 k \geq \frac {7} {12}\".<\/p>\n

Corrig\u00e9 :\u00a0<\/strong><\/p>\n

Vrai. \"\bullet\" \"S_{n + 1} = \displaystyle \sum _{k = n + 2} ^{2 n + 2} \frac 1 k\"
\n\"S_{n + 1 } = S_n \displaystyle - \frac 1 {n + 1} + \frac 1 {2 n + 1} + \frac 1 {2 n + 2}\"
\n\"S_{n + 1 } = S_n \displaystyle - \frac 1 {2 n + 2} + \frac 1 {2 n + 1}\"
\n\"S_{n + 1} - S_n = \displaystyle \frac 1 {(2 n + 1) (2 n + 2)}\"
\nLa suite \"(S_n)_{n > 1}\" est strictement croissante.<\/p>\n

\"\bullet\" \"S_2 = \displaystyle \sum _ {k = 3} ^{4} \frac 1 k = \frac 1 3 + \frac 1 4 = \frac 7 {12}\".<\/p>\n

Pour tout \"n \geq 2, \, S_n \geq S_2 \geq \displaystyle \frac {7} {12}\".<\/p>\n

Exercice 2<\/b>
\nSoit \"n \in \mathbb{N}^*\".
\n\"\displaystyle \sum _{k = 1} ^n (-1) ^{k + 1} k ^2 = (-1) ^n \, \frac {n(n + 1)} 2\"<\/p>\n

Corrig\u00e9 :<\/strong><\/p>\n

Vrai. Si \"n \in \mathbb{N}^*,\" on note\u00a0 \"H_n : \displaystyle \sum _{k = 1} ^n (-1) ^{k + 1} k ^2 = (-1) ^n \, \frac {n(n + 1)} 2\"<\/p>\n

\"\bullet\" Pour \"n = 1\"
\n\"\displaystyle S_n = \sum _{k = 1} ^1 (-1) ^{k + 1} k ^2 = - 1\" \"\displaystyle S_n = (-1) ^1 \, \frac {1(1 + 1)} 2\"
\nla propri\u00e9t\u00e9 est vraie.<\/p>\n

\"\bullet\" On suppose que \"H_n\" est vraie.
\n\"\displaystyle S_{n + 1} = \sum _{k = 1} ^{n + 1} (-1) ^{k + 1} k ^2\"
\n\"S_{n + 1} = S_n + (-1) ^{n + 1} (n + 1) ^2\"
\nEn utilisant \"H_n\" : \"S_{n + 1}\"\u00a0 est \u00e9gal \u00e0
\n\"\; = \displaystyle (-1) ^n \, \frac {n(n + 1)} 2 + (-1) ^{n + 1} (n + 1)^2\"
\n\"\; = \displaystyle (-1) ^{n + 1} \, \frac {(n + 1)} 2 \left ( 2 (n + 1) - n \right )\"
\n\"S_{n + 1} = \displaystyle (-1) ^{n + 1} \, \frac {n + 1 } 2 \left ( n + 2 \right )\"
\nce qui justifie \"H_{n + 1}\,\".<\/p>\n

La propri\u00e9t\u00e9 est d\u00e9montr\u00e9e par r\u00e9currence.<\/p>\n

Exercice 3<\/b>
\nSoit \"n \in \mathbb{N}^*\".
\n\"\displaystyle \sum _{k = 1} ^n \frac {k ^2} {4 \, k ^ 2 - 1} = \frac {n(n + 1)} {2(2 \, n + 1)}\"<\/p>\n

Corrig\u00e9 :\u00a0<\/strong><\/p>\n

Vrai. Si \"n \in \mathbb{N}^*\", on note
\n\"\quad H_n : \displaystyle \sum _{k = 1} ^n \frac {k ^2} {4 \, k ^ 2 - 1} = \frac {n(n + 1)} {2(2 \, n + 1)}\".<\/p>\n

\"\bullet\" Pour \"n = 1\",
\n\"\displaystyle S_1 = \sum _{k = 1} ^1 \frac {k ^2} {4 \, k ^ 2 - 1} = \frac 1 3 = \frac {1.2} {2 . 3}\"
\ndonc \"H_1\" est d\u00e9montr\u00e9e.<\/p>\n

\"\bullet\" On suppose que \"H_n\" est vraie.
\n\"\displaystyle S_{n + 1} = S_n + \frac {(n + 1)^2} {4(n + 1)^2 - 1}\"
\n\"\displaystyle S_{n + 1} = \frac {n(n + 1)} {2(2 \, n + 1)} + \frac {(n + 1)^2 }{(2 n + 3)(2 n + 1)}\"
\n\"\displaystyle = \frac {n + 1} {2(2 \, n + 1)(2 n + 3) }\left ( n(2 n + 3) + 2 n + 2\right )\"
\n\"n(2\, n + 3) + 2 \,n + 2 = 2 \, n ^2 + 5 \, n + 2\" \"n(2\, n + 3) + 2\, n + 2 = (2\, n + 1) (n + 2)\"
\n\"\displaystyle S_{n + 1} = \frac {(n + 1)(n + 2) } {2(2 \, n + 3) }\"
\nce qui justifie \"H_{n + 1}\,\" .<\/p>\n

La propri\u00e9t\u00e9 est d\u00e9montr\u00e9e par r\u00e9currence.<\/p>\n

Exercice 4<\/b>
\nSoit \"n \in \mathbb{N}^*\". Calculer \"S_n = \displaystyle \sum _{k = 1 } ^n \frac 1 {1 + 2 + \, \cdots \, + k} \,\".<\/p>\n

Corrig\u00e9 :\u00a0<\/strong><\/p>\n

On utilise \"\displaystyle \sum _{i = 1} ^k i = \frac {k(k + 1)} 2\"
\ndonc \"\displaystyle \frac 1 {1 + 2 + \, \cdots \, + k} = \frac 2 {k(k + 1)}\"
\npuis \"2 = 2(k + 1) - 2\, k\"
\ndonc \"\displaystyle \frac 2 {k(k + 1)} = \frac {2(k + 1) - 2 k } {k(k + 1)}\"
\n\"\displaystyle \frac 2 {k(k + 1)} = \frac {2} {k } - \frac 2 {k + 1}\".
\nOn note dans la suite \"a _k = \displaystyle \frac 2 k\".
\n\"\displaystyle \frac 2 {k(k + 1)} = a_k - a_{k + 1}\"
\n\"S_n = \displaystyle \sum _{k = 1 } ^n (a_k - a_{k + 1} )\"
\npar t\u00e9lescopage
\n\"S_n = a_1 - a_{n + 1} = \displaystyle 2 - \frac 2 {n + 1} = \frac {2 \, n} {n + 1}\,\".<\/div>\n

Exercice 5<\/b>\u00a0Formule de Vandermonde
\nSoit \"n\in \mathbb{N}^*\". \"\displaystyle \sum _{k = 0} ^n \binom n k ^2 = \binom {2n} n\".<\/p>\n

Corrig\u00e9 :\u00a0<\/strong><\/p>\n

Cette d\u00e9monstration n\u00e9cessite de savoir faire le produit de deux polyn\u00f4mes<\/a>.<\/p>\n

On utilise \"\quad (1 + \textrm{X}) ^n \, (1 + \textrm{X}) ^n = (1 + \textrm{X}) ^{2 n}\"
\nsoit \"\displaystyle \left ( \sum _{i = 0} ^n \binom {n} i \textrm{X}^i \right ) \, \left ( \sum _{j = 0} ^n \binom {n} j \textrm{X}^j \right )\" \"\displaystyle \quad \quad \quad \quad = \sum _{p = 0} ^{2\,n} \binom {2\, n} p \textrm{X}^p\".
\nPuis on \u00e9gale les coefficients de \"\textrm{X}^n\,\" :
\n\"\displaystyle \binom {2 \,n} n = \sum _{i + j = n} \binom {n} i \, \binom {n} j\"
\ndonc
\n\"\displaystyle \binom {2 \, n} n = \sum _{k = 0 }^n \binom {n} k \, \binom {n} {n - k}\"
\nce qui s’\u00e9crit aussi
\n\"\displaystyle \binom {2\, n} n = \sum _{k = 0 }^n \binom {n} k \, \binom {n} { k}\"
\net on a obtenu la relation.<\/p>\n

👍 On peut aussi d\u00e9montrer que si \"a\" et \"b \in\mathbb{N}^*\" et \"0 \leq k \leq a + b\" :
\n\"\quad \quad \displaystyle \binom {a + b } k = \sum _{ i + j = k } \binom {a} i \, \binom {b} {j}\"
\nce qui peut s’\u00e9crire
\n\"\displaystyle \binom {a + b } k = \sum _{i = \max (0 , \, k - b ) }^{\min(a , \, k)} \binom {a} i \, \binom {b} {k -i}\"<\/p>\n

Pour cela on utilise \"\quad (1 + \textrm{X}) ^a \, (1 + \textrm{X}) ^b = (1 + \textrm{X}) ^{a + b }\"
\net on d\u00e9termine le coefficient de \"\textrm{X}^k\"
\nOn transforme la somme en choisissant
\n\"\quad \; \; 0 \leq i \leq a , 0\leq k - i \leq b\"
\nssi \"0 \leq i \leq a , \, k - b \leq i \leq k \"
\nce qui donne \"\quad \quad \max (0 , \, k - b ) \leq i \leq \min(a ,\, k)\".<\/p>\n

Exercice 6<\/b>
\nSoit \"n\in \mathbb{N}^*\". Calculer\u00a0\"\displaystyle \sum _{k = 0} ^n ( - 1) ^k \binom n k ^2\"<\/p>\n

Corrig\u00e9 :\u00a0<\/strong><\/p>\n

On utilise \"(1 - \textrm{X}) ^n \, (1 + \textrm{X}) ^n = (1 - \textrm{X}^2 ) ^{ n}\"
\nsoit \"\displaystyle \left ( \sum _{i = 0} ^n \binom {n} i (- 1) ^i \textrm{X}^i \right ) \, \left ( \sum _{j = 0} ^n \binom {n} j \textrm{X}^j \right )\" \"\displaystyle \quad \quad \quad \quad = \sum _{p = 0} ^n \binom {n} p \textrm{X}^{2 \, p}\"
\net on calcule le coefficient \"c_n\" de \"\textrm{X}^n\"
\npar le produit des deux polyn\u00f4mes :
\n\"\displaystyle c_n = \sum _{i + j = n} \binom {n} i (- 1) ^i \, \binom {n} j\"
\n\"\displaystyle c_n = \sum _{k = 0}^n (- 1) ^k \binom {n} k \, \binom {n} {n - k}\"
\n\"\displaystyle c_n = \sum _{k = 0} ^n (-1) ^k \binom {n} k \, \binom {n} {k}\"<\/p>\n

Si \"n\" est impair, on en d\u00e9duit que
\n\"\displaystyle \sum _{k = 0} ^n ( - 1) ^k \binom n k ^2 = 0\"
\net si \"n = 2 \,p\" o\u00f9 \"p \in \mathbb{N}\",
\n\"\displaystyle \sum _{k = 0} ^n ( - 1) ^k \binom n k ^2 = (-1) ^p \binom {n} {p}\".<\/p>\n

Exercice 7\u00a0<\/b>
\nSoit \"n \in \mathbb{N}^*\" et \"(x_k)_{1 \leq k \leq n}\" une famille de r\u00e9els telle que
\n\"\quad \quad \quad \displaystyle \sum _ {k = 1} ^n x_k ^2 = \sum _ {k = 1} ^n x_k = n\",
\nalors \"\forall \, k \in [\! [1 , \, n]\!], x_ k = 1\"<\/p>\n

Corrig\u00e9 :\u00a0<\/strong><\/p>\n

\"\displaystyle S = \sum _ {k = 1 } ^n (x_ k - 1) ^2\"
\n\"S =\displaystyle \sum _ {k = 1} ^n x_k ^2 - 2 \sum _ {k = 1} ^n x_k + \sum _ {k = 1} ^n 1\"
\ndonc \"S = n - 2 \, n + n\".
\n\"S\" est nulle et c’est la somme de \"n\" r\u00e9els<\/a> positifs ou nuls, donc
\n\"\forall k \in [\! [1 , \, n]\!], (x_ k - 1) ^2 = 0\",\u00a0soit \"x_k = 1\".<\/p>\n

Exercice 8<\/b>
\nSi \"n \in \mathbb{N}^*, \displaystyle \sum _{k = 1}^n \binom n k \frac {(-1) ^k} k = \sum _{k = 1} ^n \frac 1 k\"<\/p>\n

Corrig\u00e9 :\u00a0<\/strong><\/p>\n

\"f(x) = \displaystyle \sum _{k = 1} ^n \binom n k \frac {(-1) ^k} k \, x^k\"
\n\"f'(x) = \displaystyle \sum _{k = 1} ^n \binom n k k \, \frac {(-1) ^k} k \, x^{k - 1}\"
\n\"f'(x) = \displaystyle \sum _{k = 1} ^n \binom n k (-1) ^k x^{k - 1}\"
\n\"x \, f'(x) = \displaystyle \sum _{k = 1} ^n \binom n k (-1) ^k x^{k}\"
\n\"x \, f'(x) = (1 - x) ^{n } - 1\".
\nSi \"x \neq 0\" , \"f'x) = - \displaystyle \frac {(1 - x) ^{n } - 1}{ (1 - x) - 1}\"
\n\"f'(x) = \displaystyle - \sum _ {k = 0} ^{n - 1} (1- x) ^{k }\"
\nLa propri\u00e9t\u00e9 reste vraie si \"x = 0\" car \"f'(0) = 0\".<\/p>\n

Puis \"f(1) - f(0) = \int_{0} ^1 f'(x) \, \textrm{d} x\"
\n\"f(1) = \displaystyle \left [ \sum _{k = 0} ^{n - 1} \frac {(1 - x)^{k + 1} }{k + 1} \right]_0 ^1\"
\n\"f(1) = - \displaystyle \sum _{k = 0} ^{n - 1}\frac 1 {k + 1}\"
\net en posant \"p = k + 1\",
\n\"\displaystyle \sum _{k = 1}^n \binom n k \frac {(-1) ^k} k = \sum _{p = 1} ^n \frac 1 n\".<\/p>\n

Exercice 9\u00a0<\/b>
\nSi \"n \in \mathbb{N}^*\", calculer \"S_n = \displaystyle \sum _ {k = 1} ^{n ^2 - 1} \left \lfloor \sqrt{k} \right \rfloor\"<\/p>\n

Corrig\u00e9 :\u00a0<\/strong><\/p>\n

On \u00e9crit une partition de \"[\! [1 \,, \, n ^2 - 1]\! ]\" faisant intervenir les carr\u00e9s des entiers entre 1 et \"n ^2 - 1\"
\n\"[\! [1 \,, \, n ^2 - 1]\! ] = \displaystyle \bigcup _{p = 1} ^{n - 1} [\! [p ^2 \,, \, (p + 1) ^2 - 1]\! ]\"
\n\"\displaystyle S_n = \sum _ {p = 1} ^{n - 1} \left ( \sum _ {k = p ^2} ^{(p + 1) ^2 - 1} \left \lfloor \sqrt{k}\right \rfloor \right )\"
\nlorsque \"k \in [\! [p ^2 \,, \, (p + 1) ^2 - 1]\! ],\, \left \lfloor \sqrt{k}\right \rfloor = p\"
\ndonc
\n\"A_p = \displaystyle \sum _ {k = p ^2} ^{(p + 1) ^2 - 1} \left \lfloor \sqrt{k} \right \rfloor = \sum _ {k = p ^2} ^{(p + 1) ^2 - 1} p\"
\n\"A_p = p\, \left ((p + 1) ^2 - p^2 \right ) = p\, (2\, p + 1 )\"
\nDonc \"S_n = \displaystyle \sum _ {k = 1} ^{n - 1} (2 \, p ^2 + p)\"
\n\"S_n = \displaystyle 2 \sum _ {p= 1 } ^{n - 1} p ^2 + \sum _ {p = 1} ^{n - 1} p\"
\n\"S_n = \displaystyle \frac {(n - 1) n (2 \, n - 1) } 3 + \frac {(n - 1) n } 2\"
\n\"S_n = \displaystyle \frac {n(n - 1)} 6 (4 \, n - 2 + 3)\"
\n\"S_n = \displaystyle \frac {n(n - 1)(4 \, n + 1)} 6\".<\/p>\n

Exercice 10<\/b>
\nCalculer si \"n \in \mathbb{N}^*\",
\n\"\quad \quad \displaystyle \sum _ {k = 1}^n \ln \left ( 1 + \frac 4 {k(k + 4)} \right )\".<\/p>\n

Corrig\u00e9 :\u00a0<\/strong><\/p>\n

\n

\"\displaystyle 1 + \frac 4 {k(k + 4) }= \frac {k ^2 + 4 \, k +4} {k ( k + 4)}\"
\ndonc \"\displaystyle \ln \left ( 1 + \frac 4 {k(k + 4)} \right )\" \"\quad\quad \quad = 2 \ln(k + 2) - \ln(k) - \ln(k + 4)\"
\nOn note \"\varphi(k) = \ln(k + 2) - \ln(k)\"
\n\"\displaystyle \ln \left ( 1 + \frac 4 {k(k + 4)} \right ) = \varphi(k) - \varphi(k + 2)\"<\/p>\n

\"\displaystyle S_n= \sum _ {k = 1}^n \ln \left ( 1 + \frac 4 {k(k + 4)} \right )\"
\n\"\displaystyle S_n =\sum _ {k = 1}^n \left (\varphi(k) - \varphi(k + 2)\right )\"
\n\"\displaystyle S_n =\sum _ {k = 1}^n \varphi(k) - \sum _ {k = 1}^n \varphi(k + 2)\"
\nen posant \"p = k + 2\",
\n\"\displaystyle S_n =\sum _ {k = 1}^n \varphi(k) - \sum _ {p = 3}^{n + 2} \varphi(p)\"
\n\"S_n = \varphi(1)+ \varphi(2) - \varphi(n+1)- \varphi(n +2)\"
\n\"S_n = \ln(3) + \ln(4) - \ln(2) - \ln(n + 3 )\" \"\quad \quad + \ln(n +1) - \ln(n + 4) + \ln(n + 2)\"
\n\"S_n = \ln(6) + \ln\left ((n + 1)(n + 2)\right )\" \"\quad \quad \quad\quad \quad \quad \quad - \, \ln \left ((n + 3)(n +4) \right )\".<\/p>\n<\/div>\n

\"\$\"\\$<\/p>\n

4. Calcul de produits<\/h3>\n

Exercice 1<\/b>
\nSi \"n \in \mathbb{N}\" et \"n \geq 2\", calculer
\n\"\quad \quad \quad \displaystyle \prod _ {k = 2} ^n \left ( 1 - \frac 1 {k ^2} \right )\".<\/p>\n

Corrig\u00e9 :\u00a0<\/strong><\/p>\n

\"\bullet\" \"P_n = \displaystyle \prod _ {k = 2} ^n \left ( 1 - \frac 1 {k ^2} \right )\"
\n\"P_n = \displaystyle \prod _ {k = 2} ^n \left ( 1 - \frac 1 {k} \right )\left ( 1 + \frac 1 {k} \right )\"
\n\"P_n = A_n \, B_n\" avec
\n\"\displaystyle A _ n = \prod _ {k = 2} ^n \left ( 1 - \frac 1 {k} \right )\"
\net \"\displaystyle B_n = \prod _ {k = 2} ^n \left ( 1 + \frac 1 {k} \right )\".<\/p>\n

\"\bullet\" \"\displaystyle A _ n = \prod _ {k = 2} ^n \frac {k - 1} {k} = \frac {a_n} {b_n}\"
\navec \"\displaystyle a_n = \prod _ {k = 2} ^n (k - 1) = \prod _ {p = 1} ^{n - 1} p = (n - 1)!\"
\n\"\displaystyle b_n = \prod _ {k = 2} ^n k =\prod_{k = 1} ^n k = n!\"
\n\"A_n = \displaystyle \frac {(n - 1)! } {n !} = \frac 1 n\".<\/p>\n

\"\bullet\" \"\displaystyle B_n = \prod _ {k = 2} ^n \left ( 1 + \frac 1 {k} \right )= \frac {c_n} {b_n}\"
\navec \"\displaystyle c_n = \prod _ {k = 2} ^n (k + 1) = \prod _ {p = 3} ^{n + 1} p = \frac {(n + 1)!} 2\"
\n\"B_n = \displaystyle \frac {(n + 1)! } {2 \, n !} = \frac {n + 1} 2\".<\/p>\n

D’o\u00f9 \"\displaystyle P_n = \frac {n + 1} {2 \, n}\,\".<\/p>\n

Exercice 2\u00a0<\/b>
\nExprimer \u00e0 l’aide factorielles
\n\"\quad \quad \quad \displaystyle \prod _{k = 1} ^{n } (2\, k + 1)\".<\/p>\n

Corrig\u00e9 :\u00a0<\/strong><\/p>\n

\n

\"\displaystyle P _n = \prod _{k = 1} ^{n } (2 \, k + 1)\".
\nOn utilise ensuite \"1 =\displaystyle \frac {2 \, k} {2\, k}\"\u00a0pour obtenir :
\n\"\displaystyle P _n = \prod _{k = 1} ^{n }\left ((2 \, k) \, (2 \, k + 1)\right ) \, \prod _{k = 1} ^n \frac 1 {2\, k}\"
\n\"u_n = \displaystyle \prod _{k = 1} ^{n }\left ( (2 \, k) \, (2\, k + 1) \right )= \prod _{p = 2} ^{2n +1 } p\" \"u_n = (2 \, n + 1)!\"
\net \"v_n = \displaystyle \prod _{k = 1} ^n {(2\, k)} = 2 ^n \prod _{k = 1} ^n k = 2 ^n \, n!\"<\/p>\n

\"P_n= \displaystyle \frac {u_n} {v_n} = \frac {(2 \, n + 1)!} {2 ^n \, n!}\".<\/p>\n<\/div>\n

C’est un calcul que vous allez souvent retrouver.
\n👍 Il faudra penser \u00e0 \u00e9crire \"\quad \quad \quad \displaystyle 1 = \frac {2 . 4 . \, \cdots \, .(2\, n)} {2 . 4 . \, \cdots \, .(2\, n)}\".<\/p>\n

Exercice 3<\/b>
\nSi \"n \in \mathbb{N}^*\" et \"x \in \mathbb{R}\"
\n\"\quad \quad P_n(x) = \displaystyle \prod_ {k = 1 }^n \left ( 1 + \frac {x} k \right)\".
\nQuestion 1\u00a0<\/b>
\nExprimer \"P_n(x)\" en fonction<\/a> de \"x,\, n\" et \"P_n(x- 1)\" lorsque \"x \neq 0\".<\/p>\n

Corrig\u00e9 :\u00a0<\/strong><\/p>\n

\"\displaystyle P_n(x) = \prod_{k = 1} ^n \frac {x + k} k\"
\ndonc \"\displaystyle P_n(x - 1 ) = \prod_{k = 1} ^n \frac {x + k- 1} k\".
\nEn posant \"p = k - 1\",
\n\"\displaystyle P_n(x - 1 ) = \prod_{p = 0} ^{n - 1} \frac {x + p} {p + 1} =x \, \prod_{p = 1} ^{n - 1} \frac {x + p} {p + 1}\"<\/p>\n

Puis on utilise \"\displaystyle \frac 1 {p + 1} = \frac 1 p \, \frac {p} {p + 1}\"
\n\"\displaystyle P_n(x - 1 ) =x \, \prod_{p = 1} ^{n - 1} \frac {x + p} {p }\, \prod_{p = 1} ^{n - 1} \frac {p} {p + 1 }\"<\/p>\n

On remarque que \"\displaystyle \prod_{p = 1} ^{n - 1} \frac {p} {p + 1 } = \frac {a_n} {b_n}\"
\navec \"\displaystyle a_n = \prod_{p = 1} ^{n - 1} {p} = (n - 1)!\"
\net \"\displaystyle b_n = \prod_{p = 1} ^{n - 1} {(p +1) } = \prod_{k = 2} ^{n} {k } = n!\"
\ndonc \"\displaystyle \prod_{p = 1} ^{n - 1} \frac {p} {p + 1 } = \frac {1} {n}\"
\net alors \"\displaystyle P_n(x - 1 ) = \frac {x} n \, \prod_{p = 1} ^{n - 1} \frac {x + p} {p }\"
\ndonc \"\displaystyle P_n(x) = \frac {x + n} n \prod_{p = 1} ^{n - 1} \frac {x + p} {p }\"<\/p>\n

\"P_n(x) = \displaystyle \frac {x + n} x \, P_n(x - 1)\".<\/p>\n

Question 2<\/b>
\nExprimer si \"p \in \mathbb{N}^*\", \"P_n(p)\" \u00e0 l’aide de coefficients du bin\u00f4me.<\/p>\n

Corrig\u00e9 :\u00a0<\/strong><\/p>\n

Si \"p \in \mathbb{N}\", \"\displaystyle P_n(p) = \frac {u_n } {v_n}\"
\navec \"v_n = \displaystyle \prod _{k = 1} ^n k = n!\"
\net \"u_n = \displaystyle \prod _{k = 1} ^n {(k + p)} = \prod _{i = p + 1} ^{p + n} i\"
\nen posant \"i = k + p\",\u00a0\"u_n = \displaystyle \frac {(p + n)!}{p!}\".<\/p>\n

\"P_n(p) = \displaystyle \frac {(p + n)!} {p!\; n! } = \binom {p + n} n\".<\/p>\n

Exercice 4\u00a0<\/b>
\nSi \"n\in \mathbb{N}^*\", simplifier\u00a0\"\displaystyle \prod _{i = 1} ^{n }\; \prod _{j = 1} ^{n } i ^j\" .<\/p>\n

Corrig\u00e9 :\u00a0<\/strong><\/p>\n

Soit \"P_n = \displaystyle \prod _{i = 1} ^{n } \; \prod _{j = 1} ^{n } i ^j\" .
\n\"\displaystyle u_ i = \prod _{j = 1} ^{n } i ^j = i ^{N}\"
\navec \"N= \displaystyle \sum _{j = 1} ^n j = \frac {n(n + 1)} 2\"
\ndonc \"u_i = i ^N\" et \"P_n = \displaystyle \prod _{i = 1} ^{n } u _i = \prod _{i = 1} ^{n } i ^N\"<\/p>\n

\"P_n = \displaystyle \left ( \prod _{i = 1} ^{n } i \right ) ^N = (n!) ^N\"
\nsoit \"P_n = (n!) ^{n(n + 1)/2} \;\".<\/p>\n

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COURS DE MATHS<\/h3>\t\t<\/div>\n\t\t\t\t<\/div>\n\t\t\t\t
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Les meilleurs professeurs particuliers<\/h4>\t\t<\/div>\n\t\t\t\t<\/div>\n\t\t\t\t
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Pour progresser et r\u00e9ussir<\/h4>\t\t<\/div>\n\t\t\t\t<\/div>\n\t\t\t\t
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\n\t\t\t\n\t\t\t\t\t\t\n\t\t\t\t\t\tCours particuliers maths<\/span>\n\t\t<\/span>\n\t\t\t\t\t<\/a>\n\t\t<\/div>\n\t\t\t\t<\/div>\n\t\t\t\t<\/div>\n\t\t\t\t
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Avis Google France <\/b>\u2605\u2605\u2605\u2605\u2605<\/b><\/span> 4,9 sur 5<\/b><\/p>\t\t\t\t\t<\/div>\n\t\t\t\t\t\t<\/div>\n\t\t\t\t<\/div>\n\t\t\t\t\t\t<\/div>\n\t\t\t\t\t<\/div>\n\t\t<\/div>\n\t\t\t\t

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\n\t\t\t\t\t\t\t\t<\/div>\n\t\t\t\t\t<\/div>\n\t\t<\/div>\n\t\t\t\t\t\t\t\t<\/div>\n\t\t\t\t\t<\/div>\n\t\t<\/section>\n\t\t\t\t\t\t\t<\/div>\n\t\t\t\t<\/div>\n\t\t\n

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5. Sommes doubles<\/h3>\n

Exercice 1\u00a0<\/b>
\nSi \"n \in \mathbb{N}^*\", calculer \"\displaystyle \sum _ {1 \leq i < j \leq n} i \, j\,\".<\/p>\n

Corrig\u00e9 :\u00a0<\/strong><\/p>\n

On note \"A = \displaystyle \sum _ {1 \leq i , j \leq n} i \, j\,\".
\n\"\bullet\" Transformation de \"A\" : on introduit
\n\"S = \displaystyle \sum _ {1 \leq i < j \leq n} i \, j\,\",\u00a0 \u00a0\"T = \displaystyle \sum _ {1 \leq j < i \leq n} i \, j\"
\net \"U = \displaystyle \sum _ {1 \leq i = j \leq n} i \, j = \sum _{i = 1} ^n i ^2 \, \,\".<\/p>\n

Par sym\u00e9trie, \"S = T\".
\nComme \"[\![1 \,,\, n]\!]^2\" est r\u00e9union des trois ensembles<\/a> 2 \u00e0 2 disjoints :
\n\"\{(i ,\, j) \in \mathbb{N}^2\, , \, 1 \leq i < j \leq n\}\",
\n\"\{(i ,\, j)\in \mathbb{N}^2 \, , \, 1 \leq j < i \leq n\}\",
\net \"\{(i , \, i)\in \mathbb{N}^2 \, , \, 1 \leq i \leq n\}\",
\n\"A = S + T + U = 2 S + U\".<\/p>\n

\"\bullet\" Calcul de \"A\".
\n\"A = \displaystyle \sum _ {1 \leq i , j \leq n} i \, j\"
\n\"A = \displaystyle \left ( \sum _ {i = 1} ^n i \right ) \left ( \sum _ {j = 1} ^n j \right )\"
\n\"A = \displaystyle \left ( \frac {n (n + 1)} 2 \right ) ^2 = \frac {n^2(n + 1)^2}4\".<\/p>\n

\"\bullet\"\u00a0Conclusion<\/b>
\n\"U = \displaystyle \sum _ {i = 1} ^n i^2 = \frac {n(n + 1)(2 n + 1)} 6\".
\n\"2 \, S = A - U\" \"2 \, S = \displaystyle \frac {n^2(n + 1)^2}4 - \frac {n(n + 1)(2\, n + 1)} 6\"
\n\"2\, S = \displaystyle \frac {n (n + 1)} {12} (3\, n ^2 + 3 \, n - 4 \, n - 2)\"
\net \"3\, n ^2 - n - 2 = (n - 1)(3\, n + 2)\"
\n\"S =\displaystyle \frac {n (n^2 - 1)(3\, n + 2)} {24}\".<\/p>\n

Exercice 2\u00a0<\/b>
\nSi \"n \in \mathbb{N}^*\", calculer \"\displaystyle \sum _ {1 \leq i \leq j \leq n} i ^2\".<\/p>\n

Corrig\u00e9 :\u00a0<\/strong><\/p>\n

\"A =\displaystyle \sum _ {1 \leq i \leq j \leq n} i ^2\"
\n\"A =\displaystyle \sum _ {i = 1} ^n \left ( \sum _ {j = i} ^n i ^2 \right )\"
\n\"A = \displaystyle \sum _ {i = 1} ^n \left (i ^2 (n - i + 1) \right )\"
\n\"A = \displaystyle (n + 1) \, \sum _ {i = 1} ^n i ^2 -\sum _ {i = 1} ^n i ^3\"
\n\"A = \displaystyle \frac {n(n + 1) ^2 (2 \, n + 1)} 6 - \frac {n^2(n + 1)^2} 4\"
\n\"A = \displaystyle \frac {n( n + 1)^2} {12} ( 4\, n + 2 - 3 \, n)\"
\n\"A = \displaystyle \frac {n( n + 1)^2(n + 2)} {12}\".<\/p>\n

Exercice 3\u00a0<\/b>
\nSi \"n \in \mathbb{N}^*\", calculer \"\displaystyle \sum _ {i = 1} ^n \sum _{j = 1} ^n \min(i \, , \, j)\".<\/p>\n

Corrig\u00e9 :\u00a0<\/strong><\/p>\n

\"A = \displaystyle \sum _ {i = 1} ^n \sum _{j = 1} ^n \min(i \, , \, j)\".<\/p>\n

\"\bullet\" On traduit d’abord si \"1 \leq i \leq n\"
\n\"S_i = \displaystyle \sum _{j = 1} ^n \min(i \, , \, j)\" \"S_i \displaystyle = \sum _{j = 1} ^i \min(i \, , \, j) + \sum _{j = i + 1} ^n \min(i \, , \, j)\"
\ndans la premi\u00e8re somme, \"\min(i , j) = j\" et dans la deuxi\u00e8me somme, \"\min(i , j) = i\",
\n\"S_i = \displaystyle \sum _{j = 1} ^i j + \sum _{j = i + 1} ^n i\"
\n\"S_i = \displaystyle \frac {i(i + 1)} 2 + (n - i)\, i\"
\n\"S_i = \displaystyle \frac {(2 n + 1) i - i ^2} 2\".<\/p>\n

\"\bullet\" Conclusion
\n\"2 \, A = \displaystyle(2 n + 1) \sum _ {i = 1} ^n i - \sum _{i = 1} ^n i ^2\"
\n\"= \displaystyle (2 n + 1) \frac {n(n + 1) } 2 - \frac {n(n + 1) (2 \, n + 1) } 6\"
\n\"2\, A = \displaystyle \frac {n(n + 1) (2\, n + 1) } 3\"
\net \"A = \displaystyle \frac {n(n + 1) (2 \, n + 1) } 6\".<\/p>\n

Exercice 4\u00a0<\/b>
\nSi \"n \in \mathbb{N}\", calculer
\n\"\quad \quad \displaystyle \sum _ {i = 0} ^n \sum _{j = 0} ^n 2 ^{\min(i \, , \, j)}\; 3 ^{\max(i , j)}\".<\/p>\n

Corrig\u00e9 :\u00a0<\/strong><\/p>\n

\"\displaystyle A = \sum _ {i = 0} ^n \sum _{j = 0} ^n 2 ^{\min(i \, , \, j)}\, 3 ^{\max(i , j)}\"
\n\"\bullet\" On traduit d’abord si \"0 \leq i \leq n\",
\n\"S_i = \displaystyle \sum _{j = 0} ^n 2 ^{\min(i \, , \, j)} \, 3 ^{\max(i , j)}\" \"S_i \displaystyle = \sum _{j = 1} ^i 2 ^{\min(i \, , \, j) } \, 3 ^{\max(i , j)}\" \"\quad \quad \quad \displaystyle + \sum _{j = i + 1} ^n 2 ^{\min(i \, , \, j)} \, 3 ^{\max(i , j)}\"
\nDans la premi\u00e8re somme, \"\min(i , j) = j\" , \"\max(i , j) = i\" et dans la deuxi\u00e8me somme, \"\min(i , j) = i\", \"\max(i , j) = j\",
\n\"S_i \displaystyle = \sum _{j = 0} ^i 2 ^{ j } \, 3 ^i + \sum _{j = i + 1} ^n 2 ^{i }\, 3 ^j\"
\nen posant \"p = j - i - 1\",
\n\"\displaystyle \sum _{j = i + 1} ^n 3 ^j =\sum _{p = 0} ^{n - i - 1} 3 ^{p + i + 1}\" \"\displaystyle \sum _{j = i + 1} ^n 3 ^j = 3 ^{i + 1} \frac {3 ^ {n - i} - 1} {3 - 1} \"
\n\"S_i \displaystyle =\left ( 2 ^{i + 1} - 1 \right ) \, 3 ^i + 2 ^i \; \frac {3 ^ {n +1} - 3 ^{i + 1} } 2\"
\n\"S_i = \displaystyle \frac 1 2 \, 6 ^i - 3 ^i + \frac 1 2 2 ^i \, 3^{n + 1}\".<\/p>\n

\"\bullet\"\u00a0Conclusion<\/b>
\n\"A = \displaystyle \sum _ {i = 0} ^n S_i\"
\n\"A = \displaystyle \frac 1 2 \sum _ {i = 0} ^n 6 ^i - \sum _ {i = 0} ^n 3 ^i + \frac {3 ^{n + 1}} 2 \sum _ {i = 0} ^n 2 ^i\"
\n\"A = \displaystyle \frac 1 2 \frac{ 6 ^{n + 1} - 1} {5} - \frac { 3 ^{n + 1} - 1} {2}\" \"\displaystyle \quad \quad \quad \quad \quad \quad + \frac {3 ^{n + 1 }} 2 \; \frac {2 ^{n + 1} - 1} 1\"
\n\"A = \displaystyle \frac 3 5 \, 6 ^{n + 1} - {3 ^{n+1}} + \frac 2 5\".<\/p>\n

Exercice 5\u00a0<\/b>
\nSi \"n \in \mathbb{N}\" et \"n \geq 2\", calculer
\n\"\quad \quad \quad \displaystyle \sum _ {i = 1} ^{n - 1} \sum _{j = i +1 } ^n \frac {i^2 } j\".<\/p>\n

Corrig\u00e9 :\u00a0<\/strong><\/p>\n

\"\bullet\" Pour intervertir les signes \"\sum\", on \u00e9crit la double somme avec un seul signe \"\sum\"
\n\"\displaystyle S = \sum _ {i = 1} ^{n - 1} \sum _{j = i + 1 } ^n \frac {1} j = \sum_ {1 \leq i < j\leq n } \frac {i^2} j\"
\nsi \"j\" est fix\u00e9 entre \"2\" et \"n\", \"i\" varie de 1 \u00e0 \"j - 1\" :
\n\"S = \displaystyle \sum _ {j = 2} ^n \sum _{i = 1} ^{j - 1} \frac {i ^2 } j = \sum _ {j = 2} ^n \frac {j(j - 1)(2 j - 1) } {6 \, j}\"
\non peut commencer la somme \u00e0 \"j = 1\" car le terme est nul si \"j = 1\"
\n\"S = \displaystyle \sum _ {j = 1} ^n \frac {(j - 1)(2 \, j - 1) } {6 }\".<\/p>\n

\"\bullet\" Fin du calcul
\n\"S = \displaystyle \frac 1 6 \left ( 2 \sum _ {j = 1} ^n j ^2 -3 \sum _ {j = 1} ^n j + n \right )\"
\n\"= \displaystyle \frac 1 6 \left ( \frac {n(n + 1)(2 n + 1)} 3 - \frac {3 \, n(n + 1) } 2 + n \right )\"
\n\"\displaystyle S = \frac {n } {36} \left ( 2\, (n + 1) (2 n + 1) - 9\, (n + 1) + 6\right )\"
\n\"S = \displaystyle \frac {n(4 n ^2 - 3 n - 1)} {36} = \frac {n(n - 1) (4 n - 1)} {36}\"<\/p>\n

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COURS PARTICULIERS EN LIGNE<\/h3>\t\t<\/div>\n\t\t\t\t<\/div>\n\t\t\t\t
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Nous avons s\u00e9lectionn\u00e9 pour vous les meilleurs professeurs particuliers.<\/h4>\t\t<\/div>\n\t\t\t\t<\/div>\n\t\t\t\t
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POUR ACC\u00c9L\u00c9RER MA PROGRESSION<\/h4>\t\t<\/div>\n\t\t\t\t<\/div>\n\t\t\t\t
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\n\t\t\t\n\t\t\t\t\t\t\n\t\t\t\t\t\tCours particuliers en ligne<\/span>\n\t\t<\/span>\n\t\t\t\t\t<\/a>\n\t\t<\/div>\n\t\t\t\t<\/div>\n\t\t\t\t<\/div>\n\t\t\t\t
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6. Formule d’inversion de Pascal<\/h3>\n

Question 1\u00a0<\/b>
\nSoit \"n \in \mathbb{N}\".
\nMontrer que pour tout \"i\in [\![0 \, , \, n]\!]\", \"\quad \displaystyle \sum _{k = i} ^n (-1) ^{n - k} \binom {n } k \binom {k} i = \binom n i \delta _{n , \, i}\"
\non note \"\delta_ {n , \, p} = 0\" si \"n \neq p\" et 1 si \"n = p\".<\/p>\n

Corrig\u00e9 :\u00a0<\/strong><\/p>\n

\"S = \displaystyle \sum _{k = i} ^n (-1) ^{n - k} \binom {n } k \binom {k} i\"
\n\"S = \displaystyle \sum _{k = i} ^n (-1) ^{n - k} \frac {n!} {k! \, (n - k)!} \, \frac {k!} {i ! \, (k - i)! }\"
\n\"S = \displaystyle \sum _{k = i} ^n \frac {(-1) ^{n - k} n!} {i! \, (n - i)!} \, \frac {(n - i)! } {(n - k) ! \, (k - i)!}\"
\n\"S = \displaystyle\binom {n} i \sum _{k = i} ^n (-1) ^{n - k} \binom {n - i} {k -i }\"<\/div>\n
puis en posant \"p = k - i\" :
\n\"S = \displaystyle\binom {n} i \sum _{p = 0} ^{n - i} (-1) ^{n - i - p} \binom {n - i} {p}\"
\n\"S = \displaystyle\binom {n} i ( 1 - 1) ^{n - i} = \binom {n} i \, \delta_{n , \, i}\"
\ncar on obtient \"1\" si \"n = i\" et \"0\" si \"n > i\".<\/div>\n

Question 2<\/b>
\nSoit \"N \in \mathbb{N}^*\" et \"(u_n)_{n \leq n \leq N}\" et \"(v_n)_{n \leq n \leq N}\" deux familles de r\u00e9els ou complexes.
\n\"\forall \, n \leq N ,\displaystyle u_n = \sum _{k = 0} ^n \binom n k v _k\"
\nssi \"\forall \, n \leq N ,\displaystyle v_n = \sum _{k = 0} ^n (-1) ^{n - k} \binom n k u _k \,\".<\/p>\n

Corrig\u00e9 :\u00a0<\/strong><\/p>\n

\"\bullet\" Condition n\u00e9cessaire.
\n\"\displaystyle T_n = \sum _{k = 0} ^n \binom n k (-1) ^{n - k} u _k\"
\nEn utilisant la d\u00e9finition de \"u_k\,\" :
\n\"\displaystyle T_n = \sum _{k = 0} ^n \binom n k (- 1) ^{n - k} \, \left ( \sum _{i = 0} ^k \binom k i v _i \right )\"
\n\"\displaystyle T_n = \sum _{k = 0} ^n \, \left ( \sum _{i = 0} ^k \binom n k (- 1) ^{n - k} \binom k i v _i \right )\"<\/p>\n

Dans le but d’intervertir les sommes :
\n\"\displaystyle T_n = \sum _{0 \leq i \leq k \leq n } (-1) ^{n - k} \binom n k \; \binom k i v _i\"
\n\"\displaystyle T_n = \sum _ {i = 0 } ^n \left ( \sum _{k = i} ^n (-1) ^{n - k} \binom n k \; \binom k i v _i \right )\"
\n\"\displaystyle T_n = \sum _ {i = 0 } ^n \left (v_i \, \sum _{k = i} ^n (-1) ^{n - k} \binom n k \; \binom k i \right )\"
\non utilise la premi\u00e8re question :
\n\"\displaystyle T_n = \sum _ {i = 0 } ^n \binom n i \; v _i \; \delta_{n ,\, i}\"
\nseul le terme pour \"i = n\" subsiste :
\n\"\displaystyle T_n = \binom n n \; v _n \; \delta_{n , \, n} = v_n\, \".
\nOn a donc \u00e9tabli la condition n\u00e9cessaire.<\/p>\n

\"\bullet\" Condition suffisante.
\nOn d\u00e9finit pour tout \"n \in [\![0 \, , \, N]\! ],\"
\n\"\quad \, u'_n = (-1) ^n \, v_n \;\" et \"v'_n = (-1) ^n \, u_n\"
\nLa premi\u00e8re partie donne puisque
\n\"\forall n \leq N ,\; \displaystyle u'_n = \sum _{k = 0} ^n \binom n k v' _k\"
\n\"\displaystyle v'_n = \sum _{k = 0} ^n (-1) ^{n - k} \binom n k u' _k\"
\nsoit apr\u00e8s multiplication par \"(-1) ^n\"
\n\"\displaystyle u_n = \sum _{k = 0} ^n (-1) ^{n - k} (-1) ^n \binom n k (-1) ^k v _k\"
\nsoit \"\displaystyle u_n = \sum _{k = 0} ^n \binom n k v _k \,\".
\nOn a\u00a0 donc prouv\u00e9 la r\u00e9ciproque.<\/p>\n

👍 Cette formule peut \u00eatre utilis\u00e9e pour calculer le nombre de surjections \"S_{p,n}\" d’un ensemble \u00e0 \"p\" \u00e9l\u00e9ments dans un ensemble \u00e0 \"m\" \u00e9l\u00e9ments ou le nombre \"D_n\" de permutations d’un ensemble \u00e0 \"n\" \u00e9l\u00e9ments sans point fixe (telles que pour tout \"x\", \"f(x) \neq x\").<\/p>\n

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Prenez de l’avance sur les cours de maths au programme de Maths Sup avec nos autres cours en ligne et exercices corrig\u00e9s sur les chapitres suivants :<\/p>\n