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adriana_98
Pe: 2013-11-09T12:01:23+02:00 In: In:

parte fractionara

Sa se demonstreze ca pentru oricare

    \[n \in N\]

, are loc egalitatea

    \[\left[ {\frac{{n + 1}}{2}} \right] + \left[ {\frac{{n + 2}}{{{2^2}}}} \right] + \left[ {\frac{{n + {2^2}}}{{{2^3}}}} \right] + .... + \left[ {\frac{{n + {2^k}}}{{{2^{k + 1}}}}} \right] + ... = n\]

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2 raspunsuri

  1. ghioknt
    profesor

    Fie n=c_p2^p+c_{p-1}2^{p-1}+c_{p-2}2^{p-2}+...+c_12+c_0 scrierea in baza 2 a lui n, unde c_i\in\{0,1\},\,c_p=1.
    Pentru k=p: [\frac{n+2^p}{2^{p+1}}]=[\frac{2^p+c_p2^p+c_{p-1}2^{p-1}+...}{2^{p+1}}=\frac{2^{p+1}+c_{p-1}2^{p-1}+...}{2^{p+1}}]=1=c_p.
    Pentru k<p: \left[\frac{n+2^k}{2^{k+1}}\right]=\left[\frac{c_p2^p+...+c_{k+1}2^{k+1}+(c_k+1)2^k+c_{k-1}2^{k-1}+...}{2^{k+1}}\right]=c_p2^{p-k-1}+...+c_{k+1}+a_k
    unde a_k este contributia la partea intreaga a fractiei \frac{c_k+1}{2} care are partea intreaga 1 daca c_k=1, 0, daca c_k=0, deci a_k=c_k.
    Sa insumam urmatoarele rezultate:
    \left[\frac{n+1}{2}\right]=c_p2^{p-1}+c_{p-1}2^{p-2}+c_{p-2}2^{p-3}+...+c_22+c_1+c_0
    \left[\frac{n+2}{2^2}\right]=c_p2^{p-2}+c_{p-1}2^{p-3}+c_{p-2}2^{p-4}+...+c_2+c_1
    \left[\frac{n+2^2}{2^3}\right]=c_p2^{p-3}+c_{p-1}2^{p-4}+c_{p-2}2^{p-5}+...+c_3+c_2
    ………
    \left[\frac{n+2^{p-1}}{2^p}\right]=c_p+c_{p-1}
    \left[\frac{n+2^p}{2^{p+1}}\right]=c_p
    Obtinem: suma din membrul I = c_p(1+2+...+2^{p-2}+2^{p-1})+c_{p-1}(1+2+...+2^{p-2})+...+c_2(1+2)+c_1+(c_p+c_{p-1}+...+c_2+c_1+c_0)=
    =c_p(2^p-1)+c_{p-1}(2^{p-1}-1)+...+c_2(2^2-1)+c_1(2-1)+(c_p+c_{p-1}+...+c_2+c_1+c_0)=
    =c_p2^p+c_{p-1}2^{p-1}+c_{p-2}2^{p-2}+...+c_12+c_0=n.
    Pentru k>p toate fractiile au partea intrega 0.
    Cu bine, ghioknt.

  2. adriana_98

    Multumesc frumos!

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