Problema Qual o valor da express\u00e3o [tex] \\boxed{1 \\cdot 2 \\cdot 3 +2 \\cdot 3 \\cdot 4 + 3 \\cdot 4 \\cdot 5 + … + 98 \\cdot 99 \\cdot 100}[\/tex]?<\/p>\n \n<\/div>\n \n Ajuda<\/strong> <\/p>\n As duas solu\u00e7\u00f5es que apresentaremos para este problema utilizam propriedades dos coeficientes binomiais que podem ser visualizadas no Tri\u00e2ngulo de Pascal<\/strong>. Elas ser\u00e3o apresentadas e ilustradas neste Lembrete<\/strong>.<\/p>\n \n Propriedade 1: <\/strong><\/span> Sejam [tex]k[\/tex] e [tex]t[\/tex] n\u00fameros naturais n\u00e3o nulos. Ent\u00e3o:<\/p>\n \n[tex] \\qquad \\qquad \\begin{pmatrix} k \\\\ k \\end{pmatrix} + \\begin{pmatrix} k + 1 \\\\ k \\end{pmatrix} + \\begin{pmatrix} k + 2 \\\\ k \\end{pmatrix} + … + \\begin{pmatrix} k + t \\\\ k \\end{pmatrix} = \\begin{pmatrix} k + t + 1 \\\\ k + 1 \\end{pmatrix} [\/tex]<\/p>\n Esta propriedade \u00e9 conhecida como Teorema das Colunas do Tri\u00e2ngulo de Pascal<\/em><\/strong>, pois ela pode ser enunciada a partir do Tri\u00e2ngulo de Pascal da seguinte forma:<\/p>\n \n Propriedade 2:<\/strong><\/span> Rela\u00e7\u00e3o de Stifel<\/strong> – Sejam [tex]k[\/tex] e [tex]t[\/tex] n\u00fameros naturais n\u00e3o nulos. Ent\u00e3o:<\/p>\n \n [tex] \\qquad \\qquad \\begin{pmatrix} k \\\\ t \\end{pmatrix} + \\begin{pmatrix} k \\\\ t+1 \\end{pmatrix} = \\begin{pmatrix} k+1 \\\\ t+1 \\end{pmatrix} [\/tex].<\/p>\n Esta propriedade tamb\u00e9m pode ser enunciada a partir do Tri\u00e2ngulo de Pascal da seguinte forma:<\/p>\n \nPara relembrar um pouco sobre coeficientes binomiais e Para [tex]n[\/tex] e [tex]r[\/tex] n\u00fameros naturais tais que [tex]0\\leqslant r \\leqslant n[\/tex], chamamos de coeficiente binomial<\/em><\/strong> ou n\u00famero binomial<\/em><\/strong> o quociente denotado por [tex]\\dbinom{n}{r}[\/tex] e assim definido: [tex]\\boxed{\\dbinom{n}{r}=\\dfrac{n!}{(n-r)! \\, r!}}[\/tex]. (Em se tratando de fatoriais, \u00e9 sempre bom se lembrar de que [tex]0!=1[\/tex].) [tex]\\textcolor{#800000}{\\dbinom{0}{0}\\\\ Observe que cada linha do Tri\u00e2ngulo de Pascal possui uma quantidade finita de elementos, mas as colunas do Tri\u00e2ngulo possuem uma quantidade infinita de elementos. \n [tex] \n (*)<\/span> Tenham em conta que a maneira diferente de escrevermos as Propriedades 1 e 2<\/strong> n\u00e3o s\u00e3o demonstra\u00e7\u00f5es e nem justificativas para elas.<\/p>\n<\/div>\n \n Solu\u00e7\u00e3o 1<\/p>\n Se [tex] x = 1\\cdot2\\cdot3 + 2\\cdot3\\cdot4 + 3\\cdot4\\cdot5 + … + 98\\cdot99\\cdot100[\/tex], ent\u00e3o: Solu\u00e7\u00e3o elaborada pelos Moderadores do Blog<\/strong>.<\/p>\n<\/div>\n \n Solu\u00e7\u00e3o 2<\/p>\n Se [tex] x = 1\\cdot2\\cdot3 + 2\\cdot3\\cdot4 + 3\\cdot4\\cdot5 + … + 98\\cdot99\\cdot100[\/tex], ent\u00e3o: Solu\u00e7\u00e3o elaborada pelos Moderadores do Blog<\/strong>.<\/p>\n<\/div>\n \n","protected":false},"excerpt":{"rendered":" Problema (Indicado a partir do 2\u00ba ano do E. M.) Qual o valor da express\u00e3o [tex] \\boxed{1 \\cdot 2 \\cdot 3 +2 \\cdot 3 \\cdot 4 + 3 \\cdot 4 \\cdot 5 + … + 98 \\cdot 99 \\cdot 100}[\/tex]? Ajuda As duas solu\u00e7\u00f5es que apresentaremos para este problema utilizam propriedades dos coeficientes binomiais que … <\/p>\n
\n (Indicado a partir do 2\u00ba ano do E. M.)<\/span><\/p>\n
\n![\"explicador_p\"](\"http:\/\/clubes.obmep.org.br\/blog\/wp-content\/uploads\/2013\/01\/menino.jpg\")
<\/p>\n\n
![\"\"](\"http:\/\/clubes.obmep.org.br\/blog\/wp-content\/uploads\/2019\/06\/pascal02.png\")
<\/p>\n\n
![\"\"](\"http:\/\/clubes.obmep.org.br\/blog\/wp-content\/uploads\/2019\/06\/pascal03.png\")
<\/p>\n
\nsobre o Tri\u00e2ngulo de Pascal, clique no bot\u00e3o abaixo.<\/strong><\/p>\n
\nNesta nossa breve discuss\u00e3o, no n\u00famero binomial [tex]\\dbinom{n}{r}[\/tex], denominaremos [tex]n[\/tex] de \u00edndice superior e [tex]r[\/tex] de \u00edndice inferior.
\nSe [tex]n[\/tex] e [tex]r[\/tex] s\u00e3o n\u00fameros naturais tais que [tex]0\\leqslant r \\leqslant n[\/tex], chamamos de Tri\u00e2ngulo de Pascal<\/em><\/strong> ao tri\u00e2ngulo num\u00e9rico infinito formado por coeficientes binomiais [tex]\\dbinom{n}{r}[\/tex] dispostos em linhas e colunas de forma que: <\/p>\n\n
\n\\dbinom{1}{0} \\dbinom{1}{1}\\\\
\n\\dbinom{2}{0} \\dbinom{2}{1} \\dbinom{2}{2}\\\\
\n\\dbinom{3}{0} \\dbinom{3}{1} \\dbinom{3}{2}\\dbinom{3}{3}\\\\
\n\\dbinom{4}{0} \\dbinom{4}{1} \\dbinom{4}{2}\\dbinom{4}{3}\\dbinom{4}{4}}\\\\
\n\\qquad \\qquad \\quad \\vdots\\\\
\n\\textcolor{#800000}{\\dbinom{n}{0} \\dbinom{n}{1} \\dbinom{n}{2}\\dbinom{n}{3} \\dbinom{n}{4} \\cdots \\dbinom{n}{n}}\\\\
\n\\qquad \\qquad \\quad \\vdots\\\\
\n[\/tex] <\/p>\n
\nSe for conveniente, podemos escrever o Tri\u00e2ngulo de Pascal substituindo cada coeficiente binomial por seu respectivo valor ou at\u00e9 mesmo dispor os n\u00fameros binomiais de forma a obter uma forma\u00e7\u00e3o diferente do Tri\u00e2ngulo.
\nVeja o Tri\u00e2ngulo de Pascal com os coeficientes binomiais que definem as sete primeiras linhas calculados e visualizado nas duas formas mais utilizadas:<\/p>\n\n
\n \n
\n\\textcolor{#800000}{1 \\, \\\\
\n1 \\, \\quad 1 \\, \\\\
\n1 \\, \\quad 2 \\, \\quad 1 \\, \\\\
\n1 \\, \\quad 3 \\, \\quad 3 \\, \\quad \\, 1 \\, \\\\
\n1 \\, \\quad 4 \\, \\quad 6 \\, \\quad \\, 4 \\, \\quad \\, 1 \\, \\\\
\n1 \\, \\quad 5 \\, \\quad 10 \\quad 10 \\quad 5 \\, \\quad \\, 1\\\\
\n1 \\, \\quad 6 \\, \\quad 15 \\quad 20 \\quad 15 \\quad 6 \\quad \\, 1\\\\
\n1 \\, \\quad 7 \\, \\quad 21 \\quad 35 \\quad 35 \\quad 21 \\quad 7 \\quad 1}\\\\
\n\\qquad \\qquad \\quad \\vdots [\/tex]<\/td>\n\\begin{equation}\\textcolor{#800000}{1 \\, \\\\
\n1 \\, \\quad \\, 1 \\, \\\\
\n1 \\, \\quad \\, 2 \\, \\quad \\, 1 \\, \\\\
\n1 \\, \\quad \\, 3 \\, \\quad \\, 3 \\, \\quad \\, 1 \\, \\\\
\n1 \\, \\quad \\, 4 \\, \\quad \\, 6 \\, \\quad \\, 4 \\, \\quad \\, 1 \\, \\\\
\n1 \\, \\quad \\, 5 \\, \\quad \\, 10 \\, \\quad \\, 10 \\, \\quad \\, 5 \\, \\quad \\, 1 \\, \\\\
\n1 \\, \\quad \\, 6 \\, \\quad \\, 15 \\, \\quad \\, 20 \\, \\quad \\, 15 \\, \\quad \\, 6 \\, \\quad \\, 1 \\, \\\\
\n1 \\, \\quad \\, 7 \\, \\quad \\, 21 \\, \\quad \\, 35 \\, \\quad \\, 35 \\, \\quad \\, 21 \\, \\quad \\, 7 \\, \\quad 1}\\\\
\n \\vdots \\end{equation} <\/td>\n<\/tr>\n<\/table>\n<\/div>\n<\/div>\n<\/div>\n
\n
\n[tex] \\qquad \\dfrac{x}{3!} = \\dfrac{1\\cdot2\\cdot3}{3!} + \\dfrac{2\\cdot3\\cdot4}{3!} + \\dfrac{3\\cdot4\\cdot5}{3!} + … + \\dfrac{98\\cdot99\\cdot100}{3!}.\\qquad \\textcolor{#800000}{(i)} [\/tex]
\nDa defini\u00e7\u00e3o de n\u00famero binomial, tem-se que [tex] \\begin{pmatrix} n \\\\ p \\end{pmatrix} = \\dfrac{n!}{(n-p)!\\cdot p!} [\/tex]; logo, da express\u00e3o [tex] \\textcolor{#800000}{(i)} [\/tex], segue que:
\n[tex]\\qquad \\dfrac{x}{3!} = \\begin{pmatrix} 3 \\\\ 3 \\end{pmatrix} + \\begin{pmatrix} 4 \\\\ 3 \\end{pmatrix} + \\begin{pmatrix} 5 \\\\ 3 \\end{pmatrix} + … + \\begin{pmatrix} 100 \\\\ 3 \\end{pmatrix}.\\qquad \\textcolor{#800000}{(ii)} [\/tex]
\nNota-se que a express\u00e3o [tex] \\textcolor{#800000}{(ii)} [\/tex] representa a soma dos [tex]98[\/tex] primeiros termos da [tex] 4\u00aa [\/tex] coluna do Tri\u00e2ngulo de Pascal; portanto, aplicando o Teorema das Colunas do Tri\u00e2ngulo de Pascal<\/strong> na express\u00e3o [tex] \\textcolor{#800000}{(ii)} [\/tex], ter-se-\u00e1 que:
\n[tex]\\qquad \\dfrac{x}{3!} = \\begin{pmatrix} 101 \\\\ 4 \\end{pmatrix} [\/tex],
\ndonde segue que:
\n[tex]\\qquad x = 3!\\cdot \\begin{pmatrix} 101 \\\\ 4 \\end{pmatrix} \\\\
\n\\qquad x = 3! \\cdot \\dfrac{101!}{97! \\cdot 4!} \\\\
\n\\qquad x = \\dfrac{101!}{97! \\cdot 4}\\\\
\n\\qquad x = \\dfrac{101\\cdot 100 \\cdot 99 \\cdot 98}{4}\\\\
\n\\qquad x = 24497550. [\/tex]
\nPelo exposto, conclu\u00edmos que [tex] \\, \\fcolorbox{black}{#eee0e5}{$1 \\cdot 2 \\cdot 3 +2 \\cdot 3 \\cdot 4 + 3 \\cdot 4 \\cdot 5 + … + 98 \\cdot 99 \\cdot 100=24 \\, 497 \\, 550$}.[\/tex]<\/p>\n
\n
\n
\n[tex] \\qquad \\dfrac{x}{3!} = \\dfrac{1\\cdot2\\cdot3}{3!} + \\dfrac{2\\cdot3\\cdot4}{3!} + \\dfrac{3\\cdot4\\cdot5}{3!} + … + \\dfrac{98\\cdot99\\cdot100}{3!}.\\qquad \\textcolor{#800000}{(i)} [\/tex]
\nDa defini\u00e7\u00e3o de n\u00famero binomial, tem-se que [tex] \\begin{pmatrix} n \\\\ p \\end{pmatrix} = \\dfrac{n!}{(n-p)!\\cdot p!} [\/tex]; logo, da express\u00e3o [tex] \\textcolor{#800000}{(i)} [\/tex], segue que:
\n[tex]\\qquad \\dfrac{x}{3!} = \\begin{pmatrix} 3 \\\\ 3 \\end{pmatrix} + \\begin{pmatrix} 4 \\\\ 3 \\end{pmatrix} + \\begin{pmatrix} 5 \\\\ 3 \\end{pmatrix} + … + \\begin{pmatrix} 100 \\\\ 3 \\end{pmatrix}.\\qquad \\textcolor{#800000}{(ii)} [\/tex]
\nPela rela\u00e7\u00e3o de Stifel, temos que
\n [tex] \\qquad \\begin{pmatrix} k \\\\ t \\end{pmatrix} = \\begin{pmatrix} k+1 \\\\ t+1 \\end{pmatrix}-\\begin{pmatrix} k \\\\ t+1 \\end{pmatrix} [\/tex],
\nassim, podemos reescrever os termos da express\u00e3o [tex] \\textcolor{#800000}{(ii)} [\/tex] nas seguintes formas:
\n[tex] \\qquad \\begin{pmatrix} 4 \\\\ 3 \\end{pmatrix} = \\begin{pmatrix} 5 \\\\ 4 \\end{pmatrix}-\\begin{pmatrix} 4 \\\\ 4 \\end{pmatrix} [\/tex]
\n[tex]\\qquad \\begin{pmatrix} 5 \\\\ 3 \\end{pmatrix} = \\begin{pmatrix} 6 \\\\ 4 \\end{pmatrix}-\\begin{pmatrix} 5 \\\\ 4 \\end{pmatrix} [\/tex]
\n[tex]\\qquad \\begin{pmatrix} 6 \\\\ 3 \\end{pmatrix} = \\begin{pmatrix} 7 \\\\ 4 \\end{pmatrix}-\\begin{pmatrix} 6 \\\\ 4 \\end{pmatrix} [\/tex]
\n[tex]\\qquad \\begin{pmatrix} 7 \\\\ 3 \\end{pmatrix} = \\begin{pmatrix} 8 \\\\ 4 \\end{pmatrix}-\\begin{pmatrix} 7 \\\\ 4 \\end{pmatrix} [\/tex]
\n[tex]\\qquad \\qquad \\vdots[\/tex]
\n[tex]\\qquad \\begin{pmatrix} 100 \\\\ 3 \\end{pmatrix}= \\begin{pmatrix} 101 \\\\ 4 \\end{pmatrix}-\\begin{pmatrix} 100 \\\\ 4 \\end{pmatrix} [\/tex].
\nSomando membro a membro as igualdades acima podemos cancelar os v\u00e1rios termos semelhantes opostos (observe que cada minuendo \u00e9 cancelado com o subtraendo da diferen\u00e7a seguinte) e obter:
\n[tex] \\qquad \\begin{pmatrix} 4 \\\\ 3 \\end{pmatrix} + \\begin{pmatrix} 5 \\\\ 3 \\end{pmatrix} + … + \\begin{pmatrix} 100 \\\\ 3 \\end{pmatrix} = \\begin{pmatrix} 101 \\\\ 4 \\end{pmatrix}-\\begin{pmatrix} 4 \\\\ 4 \\end{pmatrix} [\/tex].
\nComo [tex] \\begin{pmatrix} 4 \\\\ 4 \\end{pmatrix} = \\begin{pmatrix} 3 \\\\ 3 \\end{pmatrix} = 1 [\/tex], teremos que:
\n[tex] \\qquad \\begin{pmatrix} 3 \\\\ 3 \\end{pmatrix} + \\begin{pmatrix} 4 \\\\ 3 \\end{pmatrix} + \\begin{pmatrix} 5 \\\\ 3 \\end{pmatrix} + … + \\begin{pmatrix} 100 \\\\ 3 \\end{pmatrix} = \\begin{pmatrix} 101 \\\\ 4 \\end{pmatrix} [\/tex].
\nAssim, segue que:
\n[tex]\\qquad x = 3!\\cdot \\begin{pmatrix} 101 \\\\ 4 \\end{pmatrix} \\\\
\n\\qquad x = 3! \\cdot \\dfrac{101!}{97! \\cdot 4!} \\\\
\n\\qquad x = \\dfrac{101!}{97! \\cdot 4}\\\\
\n\\qquad x = \\dfrac{101\\cdot 100 \\cdot 99 \\cdot 98}{4}\\\\
\n\\qquad x = 24497550 [\/tex]
\ne, portanto, [tex] \\, \\fcolorbox{black}{#eee0e5}{$1 \\cdot 2 \\cdot 3 +2 \\cdot 3 \\cdot 4 + 3 \\cdot 4 \\cdot 5 + … + 98 \\cdot 99 \\cdot 100=24 \\, 497 \\, 550$} \\, .[\/tex]<\/p>\n
\n