{"id":1022,"date":"2015-01-29T09:17:07","date_gmt":"2015-01-29T08:17:07","guid":{"rendered":"http:\/\/www.unimath.fr\/?p=1022"},"modified":"2016-05-28T15:14:22","modified_gmt":"2016-05-28T14:14:22","slug":"dm-guide-suites-n2","status":"publish","type":"post","link":"http:\/\/www.unimath.fr\/?p=1022","title":{"rendered":"DM Guid\u00e9 Suites n\u00b03"},"content":{"rendered":"

Sujet :\u00a0Cliquer ici<\/a><\/strong><\/p>\n

Correction<\/strong> :\u00a0Cliquer ici<\/a><\/p>\n

    \n
  1. Calcul de u_2
    \n[peekaboo_link name=\u00a0\u00bbquestion1″]R\u00e9ponse[\/peekaboo_link][peekaboo_content name=\u00a0\u00bbquestion1″] u_2=\\frac{3}{4}[\/peekaboo_content]
    \nEn d\u00e9duire que la suite n’est pas arithm\u00e9tique
    \n[peekaboo_link name=\u00a0\u00bbquestion2″]Aide[\/peekaboo_link][peekaboo_content name=\u00a0\u00bbquestion2″]A partir des trois termes u_0, u_1, u_2, on peut d\u00e9montrer qu’elle n’est pas arithm\u00e9tique car u_2-u_1 \\neq u_1-u_0. Par contre, il ne suffit pas de regarder trois termes pour d\u00e9montrer qu’elle est arithm\u00e9tique ![\/peekaboo_content]
    \nEn d\u00e9duire que la suite n’est pas g\u00e9om\u00e9trique
    \n[peekaboo_link name=\u00a0\u00bbquestion3″]Aide[\/peekaboo_link][peekaboo_content name=\u00a0\u00bbquestion3″] M\u00eame remarque que pr\u00e9c\u00e9demment avec ici \\frac{u_2}{u_1} \\neq \\frac{u_1}{u_0}[\/peekaboo_content]<\/li>\n
  2. a. Calcul de v_0
    \n[peekaboo_link name=\u00a0\u00bbquestion3b\u00a0\u00bb]R\u00e9ponse[\/peekaboo_link][peekaboo_content name=\u00a0\u00bbquestion3b\u00a0\u00bb]v_0=1 [\/peekaboo_content]
    \nb. Expression de v_{n+1} en fonction de v_n
    \n[peekaboo_link name=\u00a0\u00bbquestion4″]Aide [\/peekaboo_link][peekaboo_content name=\u00a0\u00bbquestion4″]v_{n+1}=u_{n+2}-\\frac{1}{2}u_{n+1}[\/peekaboo_content]
    \nc. En d\u00e9duire que la suite (v_n) est g\u00e9om\u00e9trique
    \n[peekaboo_link name=\u00a0\u00bbquestion4b\u00a0\u00bb]R\u00e9ponse[\/peekaboo_link][peekaboo_content name=\u00a0\u00bbquestion4b\u00a0\u00bb] Le calcul pr\u00e9c\u00e9dent nous a donn\u00e9\u00a0v_{n+1}=\\frac{1}{2}v_n donc suite g\u00e9om\u00e9trique de raison 1\/2[\/peekaboo_content]
    \nd. Expression de v_n en fonction de n
    \n[peekaboo_link name=\u00a0\u00bbquestion4c\u00a0\u00bb]R\u00e9ponse [\/peekaboo_link][peekaboo_content name=\u00a0\u00bbquestion4c\u00a0\u00bb]v_n=v_0 \\times q^n=\\left(\\frac{1}{2}\\right)^n\u00a0[\/peekaboo_content]<\/li>\n
  3. a. Calcul de w_0
    \n[peekaboo_link name=\u00a0\u00bbquestion5″]R\u00e9ponse [\/peekaboo_link][peekaboo_content name=\u00a0\u00bbquestion5″]w_{0}=-1 [\/peekaboo_content]
    \nb. Expression de w_{n+1} en fonction de u_n et v_n
    \n[peekaboo_link name=\u00a0\u00bbquestion6″]Aide[\/peekaboo_link][peekaboo_content name=\u00a0\u00bbquestion6″] \u00a0w_{n+1}=\\frac{u_{n+1}}{v_{n+1}} avec u_{n+1}=v_n+\\frac{1}{2}u_n (d’apr\u00e8s l’\u00e9nonc\u00e9) et v_{n+1}=\\frac{1}{2}v_n[\/peekaboo_content]
    \nc. En d\u00e9duire que\u00a0w_{n+1}=w_n+2
    \n[peekaboo_link name=\u00a0\u00bbquestion8″]Aide[\/peekaboo_link][peekaboo_content name=\u00a0\u00bbquestion8″] \u00a0On peut utiliser \\frac{a+b}{c}=\\frac{a}{c}+\\frac{b}{c} pour montrer que w_{n+1}=w_n+2[\/peekaboo_content]
    \nd. Expression de w_n en fonction de n
    \n[peekaboo_link name=\u00a0\u00bbquestion9″]Aide1[\/peekaboo_link][peekaboo_content name=\u00a0\u00bbquestion9″]\u00a0(w_{n}) est arithm\u00e9tique de raison 2[\/peekaboo_content]
    \n[peekaboo_link name=\u00a0\u00bbquestion10″]Aide2[\/peekaboo_link][peekaboo_content name=\u00a0\u00bbquestion10″] donc w_{n}=w_0+nr car suite\u00a0\u00a0arithm\u00e9tique de raison 2[\/peekaboo_content]
    \n[peekaboo_link name=\u00a0\u00bbquestion11″]R\u00e9ponse[\/peekaboo_link][peekaboo_content name=\u00a0\u00bbquestion11″]\u00a0w_{n}=-1+2n [\/peekaboo_content]<\/li>\n
  4. Montrer que pour tout entier naturel n, u_n=\\frac{2n-1}{2^n}
    \n[peekaboo_link name=\u00a0\u00bbquestion12″]Aide[\/peekaboo_link][peekaboo_content name=\u00a0\u00bbquestion12″] Utiliser w_{n}=\\frac{u_n}{v_n} puis exprimer u_n en fonction de v_n et w_n (par produit en croix !) puis remplacer \u00a0 v_n et w_n par leur expression trouv\u00e9e en 2.d et 3.d[\/peekaboo_content]<\/li>\n
  5. D\u00e9monstration par r\u00e9currence
    \n[peekaboo_link name=\u00a0\u00bbquestion13″]Aide 1[\/peekaboo_link][peekaboo_content name=\u00a0\u00bbquestion13″] Exprimer\u00a0S_{n+1} en fonction de S_n pour montrer l’h\u00e9r\u00e9dit\u00e9 [\/peekaboo_content]
    \n[peekaboo_link name=\u00a0\u00bbquestion14″]Aide 2 [\/peekaboo_link][peekaboo_content name=\u00a0\u00bbquestion14″] Utiliser S_{n+1}=S_n+u_{n+1} [\/peekaboo_content]
    \n[peekaboo_link name=\u00a0\u00bbquestion15″]Aide 3[\/peekaboo_link][peekaboo_content name=\u00a0\u00bbquestion15″] Utiliser u_{n+1}=\\frac{2(n+1)-1}{2^{n+1}} [\/peekaboo_content]
    \n[peekaboo_link name=\u00a0\u00bbquestion16″]Aide 4[\/peekaboo_link][peekaboo_content name=\u00a0\u00bbquestion16″] Utiliser 2^{n+1}=2^n \\times 2 donc mettre les deux fractions sur 2^{n+1} puis r\u00e9duire en une fraction (attention aux signes !) [\/peekaboo_content]<\/li>\n<\/ol>\n","protected":false},"excerpt":{"rendered":"

    Sujet :\u00a0Cliquer ici Correction :\u00a0Cliquer ici Calcul de [peekaboo_link name=\u00a0\u00bbquestion1″]R\u00e9ponse[\/peekaboo_link][peekaboo_content name=\u00a0\u00bbquestion1″] [\/peekaboo_content] En d\u00e9duire que la suite n’est pas arithm\u00e9tique [peekaboo_link name=\u00a0\u00bbquestion2″]Aide[\/peekaboo_link][peekaboo_content name=\u00a0\u00bbquestion2″]A partir des trois termes , on peut d\u00e9montrer qu’elle n’est pas arithm\u00e9tique car . Par contre, il ne suffit pas de regarder trois termes pour d\u00e9montrer qu’elle est arithm\u00e9tique ![\/peekaboo_content] En d\u00e9duire […]<\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"closed","ping_status":"closed","sticky":false,"template":"","format":"standard","meta":[],"categories":[1],"tags":[],"_links":{"self":[{"href":"http:\/\/www.unimath.fr\/index.php?rest_route=\/wp\/v2\/posts\/1022"}],"collection":[{"href":"http:\/\/www.unimath.fr\/index.php?rest_route=\/wp\/v2\/posts"}],"about":[{"href":"http:\/\/www.unimath.fr\/index.php?rest_route=\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"http:\/\/www.unimath.fr\/index.php?rest_route=\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"http:\/\/www.unimath.fr\/index.php?rest_route=%2Fwp%2Fv2%2Fcomments&post=1022"}],"version-history":[{"count":12,"href":"http:\/\/www.unimath.fr\/index.php?rest_route=\/wp\/v2\/posts\/1022\/revisions"}],"predecessor-version":[{"id":2315,"href":"http:\/\/www.unimath.fr\/index.php?rest_route=\/wp\/v2\/posts\/1022\/revisions\/2315"}],"wp:attachment":[{"href":"http:\/\/www.unimath.fr\/index.php?rest_route=%2Fwp%2Fv2%2Fmedia&parent=1022"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"http:\/\/www.unimath.fr\/index.php?rest_route=%2Fwp%2Fv2%2Fcategories&post=1022"},{"taxonomy":"post_tag","embeddable":true,"href":"http:\/\/www.unimath.fr\/index.php?rest_route=%2Fwp%2Fv2%2Ftags&post=1022"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}