A Constant Sequence of Numbers

forum "in lucru"
Avatar utilizator
Laurențiu Ploscaru
Mesaje: 1237
Membru din: Mie Mai 04, 2011 5:42 pm
Localitate: Călimănești
Contact:

A Constant Sequence of Numbers

Mesaj de Laurențiu Ploscaru »

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)
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.
andreiteodor
Mesaje: 491
Membru din: Joi Mar 17, 2011 3:09 pm
Localitate: Sinesti

Re: A Constant Sequence of Numbers

Mesaj de andreiteodor »

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)
.
.
.
$a_3^2=6-a_2^2$ $/ * (-2)^{n-3}$
$a_2^2=6-a^1^2$ $/ * (-2)^{n-2}$
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.
andreiteodor
Mesaje: 491
Membru din: Joi Mar 17, 2011 3:09 pm
Localitate: Sinesti

Re: A Constant Sequence of Numbers

Mesaj de andreiteodor »

Aceasta problema este din shortlist onm , 2002 , clasa a X-a.
Nimic nu-i niciodata asa de simplu cum pare.
Scrie răspuns