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UncleSam
Pe: 2013-07-14T11:01:36+03:00 In: In:

2 probleme de combinatorica

1. Fie f:D \to R\], f(x) = C_{3x + 7}^{6x + 2}\], unde D este domeniul maxim de definitie. Fie M ={\max }\limits_{x \in D} f(x)\]. Atunci:

a) M=21;
b) M=84;
c) M=45;
d) M=72;
e) M=210;
f) M=60;

2. Se considera suma S = \frac{{C_n^0}}{1} + \frac{{C_n^1}}{2} + \frac{{C_n^2}}{3} + ... + \frac{{C_n^n}}{{n + 1}}\]. Avem:

a) S = {2^{n + 1}}\];
b) S = \frac{{{2^n} - 1}}{n}\];
c) S = \frac{{{2^{n + 1}} - 1}}{{n + 1}}\];
d) S = \frac{{{2^n} - 1}}{{n + 1}}\];
e) S = \frac{{{2^{n + 1}} - 1}}{n}\];
f) S = n \cdot {2^{n + 1}}\];

Similare

3 raspunsuri

  1. gunty
    maestru (V)

    1.

        \[ C_{3x + 7}^{6x + 2} = \frac{{\left( {3x + 7} \right)!}}{{\left( {6x + 2} \right)!\left( { - 3x + 5} \right)!}} \Rightarrow \left. \begin{array}{l} 3x + 7 \in N \Rightarrow x \ge - \frac{7}{3} \\ 6x + 2 \in N \Rightarrow x \ge - \frac{1}{3} \\ - 3x + 5 \in N \Rightarrow x \le \frac{5}{3} \\ x = \frac{k}{3}\;,k \in Z \\ \end{array} \right\} \Rightarrow x \in \left\{ { - \frac{1}{3},0,\frac{1}{3},\frac{2}{3},\frac{3}{3},\frac{4}{3},\frac{5}{3}} \right\}. \]

    In continuare calculezi f(x) pentru fiecare valoare din multimea primita pentru a afla max(f(x)).

    2.

        \[ \begin{array}{l} S = \frac{{C_n^0 }}{1} + \frac{{C_n^1 }}{2} + \frac{{C_n^2 }}{3} + ... + \frac{{C_n^n }}{{n + 1}} = \sum\limits_{k = 0}^n {\frac{{C_n^k }}{{k + 1}}} = \sum\limits_{k = 0}^n {\frac{{C_{n + 1}^{k + 1} }}{{n + 1}}} = \frac{1}{{n + 1}}\sum\limits_{k = 0}^n {C_{n + 1}^{k + 1} } = \frac{1}{{n + 1}}\left( {C_{n + 1}^1 + C_{n + 1}^2 + ... + C_{n + 1}^n + C_{n + 1}^{n + 1} } \right) = \\ = \frac{1}{{n + 1}}\left( {C_{n + 1}^0 + C_{n + 1}^1 + ... + C_{n + 1}^n + C_{n + 1}^{n + 1} - 1} \right) = \frac{1}{{n + 1}}\left( {2^{n + 1} - 1} \right) = \frac{{2^{n + 1} - 1}}{{n + 1}}. \\ \end{array} \]

    * Am folosit formula

        \[ \frac{{C_{n + 1}^{k + 1} }}{{n + 1}} = \frac{{C_n^k }}{{k + 1}}\;,\forall n,k \in N. \]

    EDIT: Am avut o mica greseala 😀…acum e in regula.

  2. UncleSam

    gunty wrote: 1.

        \[ C_{3x + 7}^{6x + 2} = \frac{{\left( {3x + 7} \right)!}}{{\left( {6x + 2} \right)!\left( { - 3x + 5} \right)!}} \Rightarrow \left. \begin{array}{l} 3x + 7 \in N \Rightarrow x \ge - \frac{7}{3} \\ 6x + 2 \in N \Rightarrow x \ge - \frac{1}{3} \\ - 3x + 5 \in N \Rightarrow x \le \frac{5}{3} \\ x = \frac{k}{3}\;,k \in Z \\ \end{array} \right\} \Rightarrow x \in \left\{ { - \frac{1}{3},0,\frac{1}{3},\frac{2}{3},\frac{3}{3},\frac{4}{3},\frac{5}{3}} \right\}. \]

    In continuare calculezi f(x) pentru fiecare valoare din multimea primita pentru a afla max(f(x)).

    2.

        \[ \begin{array}{l} S = \frac{{C_n^0 }}{1} + \frac{{C_n^1 }}{2} + \frac{{C_n^2 }}{3} + ... + \frac{{C_n^n }}{{n + 1}} = \sum\limits_{k = 0}^n {\frac{{C_n^k }}{{k + 1}}} = \sum\limits_{k = 0}^n {\frac{{C_{n + 1}^{k + 1} }}{{n + 1}}} = \frac{1}{{n + 1}}\sum\limits_{k = 0}^n {C_{n + 1}^{k + 1} } = \frac{1}{{n + 1}}\left( {C_{n + 1}^1 + C_{n + 1}^2 + ... + C_{n + 1}^n + C_{n + 1}^{n + 1} } \right) = \\ = \frac{1}{{n + 1}}\left( {C_{n + 1}^0 + C_{n + 1}^2 + ... + C_{n + 1}^n + C_{n + 1}^{n + 1} - 1} \right) = \frac{1}{{n + 1}}\left( {2^{n + 1} - 1} \right) = \frac{{2^{n + 1} - 1}}{{n + 1}}. \\ \end{array} \]

    * Am folosit formula

        \[ \frac{{C_{n + 1}^{k + 1} }}{{n + 1}} = \frac{{C_n^k }}{{k + 1}}\;,\forall n,k \in N. \]

    Am inteles. Multumesc mult 😀

  4. TheodorMunteanu

    consideram functia f(x)=(1+x)^n
    pe de o parte \int_0^1 f(x)dx=\frac{(x+1)^{n+1}}{n+1}|_0^1=\frac{2^{n+1}-1}{n+1}
    pe de alta parte (1+x)^n=1+C_n^1x+C_n^2 x^2+...+C_n^nx^n
    integrand aceasta de la 0 la 1 obtinem ca \int_0^1(1+C_n^1x+C_n^2x^2+...+C_n^nx^n)=(xC_n^0+\frac{x^2}{2}C_n^1+...+\frac{x^{n+1}}{n+1}C_n^n\right)|_0^1 =C_n^0+\frac{1}{2}C_n^1+...+\frac{1}{n+1}C_n^n
    de unde rezulta identitatea ceruta

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