Question

Bună seara
Se considera sirul (xn), definit prin:

xn+1-axn+2=0 xo=a


1.Multimea valorilor parametrului a pentru care sirul (xn) este strict descrescator este:

Answer 1

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Answer 2

$\displaystyle x_{n+1}-ax_n+2=0 ~~~(*)\\ \\ x_{n+2}-ax_{n+1}+2=0 ~~~(**) \\ \\ ---------- \\ \\ x_{n+2}-x_{n+1}-a(x_{n+1}-x_n)=0~~~(**)-(*) \\ \\ Notam~d_n=x_{n+1}-x_n. ~Trebuie~sa~avem~d_n<0~\forall~n \in \mathbb{N}. \\ \\ Avem~d_{n+1}=ad_n,~deci~d_n=a^nd_0. \\ \\ Evident~nu~putem~avea~a=0. \\ \\ De~asemenea~nu~putem~avea~a<0.~(in~sirul~d_n~ar~alterna~semnele). \\ \\ Rezulta~a>0,~si~cum~d_n<0~\forall~n \in \mathbb{N}~rezulta~d_0<0. \\ \\ Conditiile~a>0~si~d_0<0~sunt~necesare~si~suficiente.$

$\displaystyle d_0<0 \Leftrightarrow x_1<x_0 \Leftrightarrow ax_0-2<x_0 \Leftrightarrow a^2-2<a \Leftrightarrow a^2-a-2<0 \\ \\ \Leftrightarrow (a+1)(a-2)<0,~si~cum~a>0,~avem~a<2. \\ \\ Deci~a \in (0,2).$