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grapefruit
maestru (V)
Pe: 2014-05-04T13:38:48+03:00 In: In:

derivata

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Problema zice sa o rezolv cu ajutorul derivatelor,eu m-am gandit ca daca iau suma

(x^2)/2+(4x^3)/3+…+(n^2 *x^(n+1))/(n+1) asta conduce prin derivare la suma din enunt,iar daca gasesc o metoda de calcul al acestei sume voi deriva acea formula si voi rezolva problema.
-nu stiu sa calculez suma
-poate exista alta idee.

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  1. gunty
    maestru (V)

        \[ \begin{array}{l} A = x^3 + x^4 + ... + x^{n + 2} = \frac{{x^{n + 3} - x^3 }}{{x - 1}} \\ B = x^2 + 2x^3 + ... + nx^{n + 1} = ? \\ C = x + 2^2 x^2 + ... + n^2 x^n = ? \\ A' = 3x^2 + 4x^3 + ... + \left( {n + 2} \right)x^{n + 1} = B + 2A + 2x^2 - 2x^{n + 2} = B + \frac{{2x^{n + 3} - 2x^3 }}{{x - 1}} + 2x^2 - 2x^{n + 2} = B + 2\frac{{x^{n + 2} - x^2 }}{{x - 1}} \\ \Leftrightarrow B = A' - 2\frac{{x^{n + 2} - x^2 }}{{x - 1}} = \frac{{\left( {\left( {n + 3} \right)x^{n + 2} - 3x^2 } \right)\left( {x - 1} \right) - \left( {x^{n + 3} - x^3 } \right)}}{{\left( {x - 1} \right)^2 }} - 2\frac{{x^{n + 2} - x^2 }}{{x - 1}} = \frac{{\left( {n + 3} \right)x^{n + 2} - 3x^2 }}{{x - 1}} - \frac{{2x^{n + 2} - 2x^2 }}{{x - 1}} - \frac{{x^{n + 3} - x^3 }}{{\left( {x - 1} \right)^2 }} = \frac{{\left( {n + 1} \right)x^{n + 2} - x^2 }}{{x - 1}} - \frac{{x^{n + 3} - x^3 }}{{\left( {x - 1} \right)^2 }}. \\ \\ B' = 2x + 3 \cdot 2x^3 + ... + \left( {n + 1} \right)nx^n = \sum\limits_{k = 1}^n {k^2 x^k } + \sum\limits_{k = 1}^n {kx^k } = C + \frac{B}{x} \\ \Leftrightarrow C = B' - \frac{B}{x} = \frac{{\left[ {\left( {n + 1} \right)\left( {n + 2} \right)x^{n + 1} - 2x} \right]\left( {x - 1} \right) - \left[ {\left( {n + 1} \right)x^{n + 2} - x^2 } \right]}}{{\left( {x - 1} \right)^2 }} - \frac{{\left[ {\left( {n + 3} \right)x^{n + 2} - 3x^2 } \right]\left( {x - 1} \right)^2 - 2\left( {x^{n + 3} - x^3 } \right)\left( {x - 1} \right)}}{{\left( {x - 1} \right)^4 }} - \frac{{\left( {n + 1} \right)x^{n + 1} - x}}{{x - 1}} + \frac{{x^{n + 2} - x^2 }}{{\left( {x - 1} \right)^2 }}. \\ \\ Dupa\;calcule\;se\;obtine\;C = \frac{{2\left( {x^{n + 3} - x^3 } \right)}}{{\left( {x - 1} \right)^3 }} - \frac{{\left( {2n + 3} \right)x^{n + 2} - 3x^2 }}{{\left( {x - 1} \right)^2 }} + \frac{{\left( {n + 1} \right)^2 x^{n + 1} - x}}{{x - 1}}. \\ \end{array} \]

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