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Problema \n<\/div>\n \n Solu\u00e7\u00e3o <\/p>\n \n Solu\u00e7\u00e3o elaborada pelos Moderadores do Blog<\/strong>.<\/p>\n<\/div>\n \n","protected":false},"excerpt":{"rendered":" Problema (Indicado a partir do 8\u00ba ano do E. F.) Calcule a seguinte soma [tex]\\qquad \\qquad \\, \\, \\, \\, \\, \\, \\, \\, \\, \\, S = \\dfrac{1}{1 \\cdot 4} + \\dfrac{1}{4 \\cdot 7} + \\dfrac{1}{7 \\cdot 10} + \\ldots + \\dfrac{1}{2998 \\cdot 3001}[\/tex]. Solu\u00e7\u00e3o Perceba inicialmente que a diferen\u00e7a entre os dois fatores … <\/p>\n
\n (Indicado a partir do 8\u00ba ano do E. F.)<\/span><\/p>\n
\nCalcule a seguinte soma
\n[tex]\\qquad \\qquad \\, \\, \\, \\, \\, \\, \\, \\, \\, \\, S = \\dfrac{1}{1 \\cdot 4} + \\dfrac{1}{4 \\cdot 7} + \\dfrac{1}{7 \\cdot 10} + \\ldots + \\dfrac{1}{2998 \\cdot 3001}[\/tex].<\/p>\n
\nPerceba inicialmente que a diferen\u00e7a entre os dois fatores no denominador de cada fra\u00e7\u00e3o \u00e9 [tex]3[\/tex].
\nAl\u00e9m disso, para [tex]a \\neq 0[\/tex] e [tex]a \\neq -3[\/tex], temos
\n[tex]\\qquad \\dfrac{1}{a} – \\dfrac{1}{a+3} = \\dfrac{a +3 – a}{a \\cdot (a+3)} = \\dfrac{3}{a \\cdot (a+3)}[\/tex]
\nLogo,
\n[tex]\\qquad \\dfrac{1}{a \\cdot (a+3)} = \\dfrac{\\dfrac{1}{a} – \\dfrac{1}{a+3}}{3}, \\quad \\forall a \\in \\mathbb{R}[\/tex], [tex]a \\neq 0[\/tex] e [tex]a \\neq -3[\/tex].
\nUtilizando esse resultado em cada parcela, podemos reescrever a express\u00e3o inicial de [tex]S[\/tex] como segue:
\n[tex]\\qquad S = \\dfrac{\\dfrac{1}{1} – \\dfrac{1}{4}}{3} + \\dfrac{\\dfrac{1}{4} – \\dfrac{1}{7}}{3} + \\dfrac{\\dfrac{1}{7} – \\dfrac{1}{10}}{3} + \\ldots + \\dfrac{\\dfrac{1}{2998} – \\dfrac{1}{3001}}{3}[\/tex]
\n[tex]\\qquad S = \\dfrac{1}{3} \\cdot \\left(\\dfrac{1}{1} – \\dfrac{1}{4} + \\dfrac{1}{4} – \\dfrac{1}{7} + \\dfrac{1}{7} – \\dfrac{1}{10} + \\ldots +\\dfrac{1}{2995} – \\dfrac{1}{2998} + \\dfrac{1}{2998} – \\dfrac{1}{3001} \\right)[\/tex]
\n[tex] \\qquad S = \\dfrac{1}{3} \\cdot \\left(\\dfrac{1}{1} – \\cancel{\\dfrac{1}{4}} + \\cancel{\\dfrac{1}{4}} – \\cancel{\\dfrac{1}{7}} + \\cancel{\\dfrac{1}{7}} – \\cancel{\\dfrac{1}{10}} + \\ldots\\\\
\n\\qquad \\qquad \\quad \\quad \\;\\;\\ldots + \\cancel{\\dfrac{1}{2995}} – \\cancel{\\dfrac{1}{2998}} + \\cancel{\\dfrac{1}{2998}} – \\dfrac{1}{3001} \\right)[\/tex]
\n[tex] \\qquad S = \\dfrac{1}{3} \\cdot \\left(\\dfrac{1}{1} – \\dfrac{1}{3001} \\right)[\/tex]
\n[tex]\\qquad S = \\dfrac{1}{3}\\cdot \\left(\\dfrac{3001 – 1}{3001}\\right) [\/tex]
\n[tex]\\qquad S = \\dfrac{1}{3}\\cdot \\left(\\dfrac{3000}{3001}\\right)[\/tex]
\n[tex]\\qquad S = \\dfrac{1000}{3001}[\/tex].<\/p>\n
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