Proof that the sequence of positive numbers $(a_n)_{n\ge 1}$, which satisfies the relation: $a_{n+1}=\sqrt{6-2a_n^2}$, is constantly.
(Laurențiu Panaitopol)
A Constant Sequence of Numbers
- Laurențiu Ploscaru
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A Constant Sequence of Numbers
People are strange when you're a stranger,
Faces look ugly when you're alone.
Women seem wicked when you're unwanted,
Streets are uneven when you're down.
Faces look ugly when you're alone.
Women seem wicked when you're unwanted,
Streets are uneven when you're down.
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- Mesaje: 491
- Membru din: Joi Mar 17, 2011 3:09 pm
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Re: A Constant Sequence of Numbers
Avem $a_{n+1}^2=6-2a_n^2$ si apoi :
$a_n^2=6-2a{n-1}^2$
$a^{n-1}^2=6-2a_{n-2}^2$ /* (-2)
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$a_3^2=6-a_2^2$ $/ * (-2)^{n-3}$
$a_2^2=6-a^1^2$ $/ * (-2)^{n-2}$
Adunanad aceste relatii => $a_n=6(1+(-2)+(-2)^2+...+(-2)^{n-2})+(-2)^{n-1}a_1^2$$(-2)^n+2+(-2)^{n-1}a_1^2=(-2)^{n-1}(a_1^2-2)+2$.
Daca $a_n^2>2$ , pentru n suficient de mare $a_{2n+1}^2=-2^{2n+1}(a_1^2-2)+2<0$, imposibil. Analog contradicitie pentru $a_n^2<2$. Deci $a_n^2=2 => a_n=\sqrt{2}$.
$a_n^2=6-2a{n-1}^2$
$a^{n-1}^2=6-2a_{n-2}^2$ /* (-2)
.
.
.
$a_3^2=6-a_2^2$ $/ * (-2)^{n-3}$
$a_2^2=6-a^1^2$ $/ * (-2)^{n-2}$
Adunanad aceste relatii => $a_n=6(1+(-2)+(-2)^2+...+(-2)^{n-2})+(-2)^{n-1}a_1^2$$(-2)^n+2+(-2)^{n-1}a_1^2=(-2)^{n-1}(a_1^2-2)+2$.
Daca $a_n^2>2$ , pentru n suficient de mare $a_{2n+1}^2=-2^{2n+1}(a_1^2-2)+2<0$, imposibil. Analog contradicitie pentru $a_n^2<2$. Deci $a_n^2=2 => a_n=\sqrt{2}$.
Nimic nu-i niciodata asa de simplu cum pare.
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- Mesaje: 491
- Membru din: Joi Mar 17, 2011 3:09 pm
- Localitate: Sinesti
Re: A Constant Sequence of Numbers
Aceasta problema este din shortlist onm , 2002 , clasa a X-a.
Nimic nu-i niciodata asa de simplu cum pare.