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ioanacnpr
user (0)
Pe: 2010-10-13T12:46:31+03:00 In: In:

Sa se calculeze

n 1 a^k
S ( a^2k a^3k) unde a^2+a+1=0
k=1

Suma de la k=1 pana la n din matricea de mai sus stiind ca a^2……….

ecuatia se rezolva cu nr complexe? Si daca da,cum? Am gasit undeva ca din ecuatie rezulta a^3=1 …de unde?

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2 raspunsuri

  1. sindex
    user (0)

    Simplu, inmultesti ecuatia cu a si faci un mic artificiu:

        \[\begin{array}{l} {a^2} + a + 1 = 0|( \cdot a) \Rightarrow {a^3} + {a^2} + a = 0 \\ {a^3} + {a^2} + a = 0 \Leftrightarrow {a^3} + \underbrace {{a^2} + a + 1}_0 - 1 = 0 \Rightarrow {a^3} + 0 - 1 = 0 \Rightarrow {a^3} = 1 \\ \end{array}\]

        $${ \sum\limits_{k = 1}^n {\left( {\matrix{ 1 & {{a^k}} \cr {{a^{2k}}} & {{a^3}^k} \cr } } \right)} = \left( {\matrix{ {\sum\limits_1^n 1 } & {\sum\limits_1^n {{a^k}} } \cr {\sum\limits_1^n {{a^{2k}}} } & {\sum\limits_1^n {{a^{3k}}} } \cr } } \right) \cr (1){a_{11}} = \sum\limits_1^n 1 = n \cr (2){a_{12}} = \sum\limits_1^n {{a^k}} = \underbrace {prog.geom.}_{q = a} = \underbrace {{a^1} + {a^2} + ... + {a^n}}_{{\rm{aplici formula standard (o ai in caiet)}}} \cr (3){a_{21}} = \sum\limits_1^n {{a^{2k}}} = \underbrace {prog.geom.}_{q = {a^2}} = idem(2) \cr \left. \matrix{ (4){a_{22}} = \sum\limits_1^n {{a^{3k}}} \hfill \cr {a^3} = 1 \hfill \cr} \right| \Rightarrow \sum\limits_1^n {{a^{3k}}} = \sum\limits_1^n {{{({a^3})}^k} = \sum\limits_1^n {{1^k} = 1 + {1^2} + ... + {1^n}} } = n \cr {\rm{deci }}\sum\limits_{k = 1}^n {\left( {\matrix{ 1 & {{a^k}} \cr {{a^{2k}}} & {{a^3}^k} \cr } } \right)} = \left( {\matrix{ n & {{a_{12}}} \cr {{a_{21}}} & n \cr } } \right),{\rm{la }}{{\rm{a}}_{12}}{\rm{ si }}{{\rm{a}}_{21}}{\rm{ ai de aplicat o sg}}{\rm{. formula}}{\rm{, cea a progresiei geometrice}} \cr} $$

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  2. sindex
    user (0)

    pe

        $$\sum\limits_1^n {{a^{2k}}} $$

    o mai poti scrie :

        $$\left. \matrix{ {a^{2k}} = {{{a^{3k}}} \over {{a^k}}} \hfill \cr {a^3} = 1 \hfill \cr} \right| \Rightarrow {a^{2k}} = {{{{({a^3})}^k}} \over {{a^k}}} = {{{1^k}} \over {{a^k}}} \Rightarrow \sum\limits_1^n {{a^{2k}}} = \sum\limits_1^n {{{{1^k}} \over {{a^k}}} = {1 \over a}} + {1 \over {{a^2}}} + ... + {1 \over {{a^n}}} = prog.geom{\rm{ de ratie q = }}{1 \over a},{\rm{ unde aplici aceeasi formula de baza}}$$

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