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dona01
Pe: 2015-06-14T07:02:24+03:00 In: In:

Algebra

Calculati suma

1/3^2+2/15^2+3/35^2+4/63^2+…+ 50/9999^2

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2 raspunsuri

  1. adrianS

    Buna seara
    Avem de calculat urmatoarea suma: \dfrac{1}{3^2}+\dfrac{2}{15^2}+\dfrac{3}{35^2}+\dfrac{4}{63^2}+\dots\dfrac{50}{9999^2}
    Suma o rescriem astfel:
    \dfrac{1}{1^2\cdot3^2}+\dfrac{2}{3^2\cdot5^2}+\dfrac{3}{5^2\cdot7^2}+\dfrac{4}{7^2\cdot9^2}+\dots\dfrac{50}{9999^2}Observam urmatoarea succesiune:numitorul este fomat dintr-un produs de patratul a doua numere care difera cu doi intre ele de ex.15^2=3^2\cdot5^2=3^2\cdot(3+2)^2Aceasta ne facem sa scriem ultimul termen astfel:n(n+2)=9999.sau
    n^2+2n-9999=0o ecuatie de gradul doi cu solutia valabila n=99.
    Deci 9999=99×101.
    Deci sirul va fi:\dfrac{1}{1\cdot9}+\dfrac{2}{9\cdot25}+\dfrac{3}{25\cdot49}+\dfrac{4}{49\cdot81}+\dots\dfrac{50}{99^2\cdot101^2} Grupele se rescriu astfel:
    \dfrac{1}{1\cdot9}=0+\dfrac{1}{9};\ \ \dfrac{2}{9\cdot25}=-\dfrac{1}{9}+\dfrac{3}{25};\ \ \dfrac{3}{25\cdot49}=-\dfrac{3}{25}+\dfrac{6}{49};\ \ \dfrac{4}{49\cdot81}=-\dfrac{6}{49}+\dfrac{10}{81}etcSirul devine deci:
    0+\dfrac{1}{9}-\dfrac{1}{9}+\dfrac{3}{25}-\dfrac{3}{25}+\dfrac{6}{49}-\dfrac{6}{49}+\dfrac{10}{81}-\dfrac{10}{81}+\dfrac{15}{121}+\dots-\dfrac{1225}{99^2}+\dfrac{1275}{101^2}Se observa ca in suma se reduc succesiv termenii si ramane doar un singur termen egal cu \dfrac{1275}{101^2}
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  2. dona01

    Multumesc mult! Destul de complicat, nu m-as fi gandit!

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