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jklR7
Pe: 2018-04-11T13:03:58+03:00 In: In:

140/UTCN2018

Am încercat și reîncercat si nu ajung la nimic concret. Mă puteți ajuta cu o idee de rezolvare?

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3 raspunsuri

  1. ghioknt
    profesor

    Doar o idee: S=\sum_{k=1}^{n}\frac{x_k-1+1}{x_k-1}=\sum_{k=1}^{n}1-\sum_{k=1}^{n}\frac{1}{1-x_k}
    Presupun că știi ambele sume.

  2. Felixx
    veteran (III)

    Fie P\left ( X \right )=X^{n}-3X^{n-1}+2X+1.Atunci {P}'\left ( X \right )=nX^{n-1}-3\left ( n-1 \right )X^{n-2}+2.
    Daca x_{1},x_{2},...,x_{n} sunt radacinile polinomului, atunci P\left ( X \right )=\left ( X-x_{1} \right )\left ( X-x_{2} \right )\cdot ...\cdot \left ( X-x_{n} \right ) si {P}'\left ( X \right )=\left ( X-x_{2} \right )\left ( X-x_{3} \right )\cdot ...\cdot \left ( X-x_{n} \right )+\left ( X-x_{1} \right )\left ( X-x_{3} \right )\cdot ...\cdot \left ( X-x_{n} \right )+...+\left ( X-x_{1} \right )\left ( X-x_{2} \right )\cdot ...\cdot \left ( X-X_{n-1} \right ).Atunci \frac{{P\left ( X \right )}'}{P\left ( X \right )}=\frac{1}{X-x_{1}}+\frac{1}{X-x_{2}}+...+\frac{1}{X-x_{n}} unde pentru x=1 obtinem:
    \frac{{P}'\left ( 1 \right )}{P\left ( 1 \right )}=\frac{1}{1-x_{1}}+\frac{1}{1-x_{2}}+...+\frac{1}{1-x_{n}}=S_{2} iar {P}'\left ( 1 \right )=n-3\left ( n-1 \right )+2=-2n+5 si P\left ( 1 \right )=1-3+2+1=1.Deci S_{2}=-2n+5 si S_{1}=n.
    Prin urmare S=\sum_{k=1}^{n}\frac{x_{k}-1+1}{x_{k}-1}=\sum_{k=1}^{n}\frac{x_{k}-1}{x_{k}-1}-\sum_{k=1}^{n}\frac{1}{1-x_{k}}=\sum_{k=1}^{n}1-\sum_{k=1}^{n}\frac{1}{1-x_{k}}=S_{1}-S_{2}=n-\left ( -2n+5 \right )=3n-5
    Raspuns A)

  4. jklR7

    Vă mulțumesc!

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