Răspuns:
Explicație pas cu pas:
[tex]x\circ x=3(x+1)^2-1\\x\circ x\circ x=3{[3(x+1)^2-1+1]}(x+1)-1=9(x+1)^3-1\\\texttt{Printr-un rationam.ent inductiv rezulta ca:}\\\underbrace{x\circ x\circ x\circ\ldots\circ x}=3^{n-1}(x+1)^n-1,\forall n\in \mathbb{N},n\geq 2\\~~~~~~~\texttt{de n ori}\\\texttt{/* te las pe tine sa demonstrezi asta */}\\\underbrace{x\circ x\circ x\circ\ldots\circ x}=-\dfrac{2}{3}\\~~~~~\texttt{de 2016 ori}\\3^{2015}(x+1)^{2016}-1=-\dfrac{2}{3}\\3^{2015}(x+1)^{2016}=\dfrac{1}{3}[/tex]
[tex](x+1)^{2016}=\left(\dfrac{1}{3}\right)^{2016}\\|x+1|=\dfrac{1}{3}\\i)x+1=\dfrac{1}{3}\Rightarrow x_1=-\dfrac{2}{3}\\ii)x+1=-\dfrac{1}{3}\Rightarrow x_2=-\dfrac{4}{3}\\S:x\in\left\{\dfrac{2}{3},-\dfrac{4}{3}\right\}[/tex]
[tex]\texttt{Demonstratia prin inductie:}\\P(n): \underbrace{x\circ x\circ x\circ \ldots\circ x}=3^{n-1}(x+1)^n-1,\forall~ n\in\mathbb{N},n\geq 2\\~~~~~~~~~~~~~~~\texttt{de n ori}\\\texttt{Etapa 1. Verificarea}\\P(2): x\circ x=3(x+1)^2-1\\3(x+1)^2-1=3(x+1)^2-1,\texttt{ adevarat.}\\\\\texttt{Etapa 2. Demonstratia propriu-zisa}\\\texttt{Presupunem P(k) adevarat }\forall k\in\mathbb{N},k\geq 2.\texttt{ Se demonstreaza}\\\texttt{ ca si P(k+1) este adevarat.}[/tex]
[tex]P(k):\underbrace{x\circ x\circ x\circ\ldots\circ x}=3^{k-1}(x+1)^k-1,~\forall k\in\mathbb{N},k\geq 2\\~~~~~~~~~~~~~~~\texttt{de k ori}\\ P(k+1):\underbrace{x\circ x\circ x\circ\ldots\circ x}=3^{k}(x+1)^{k+1}-1\\~~~~~~~~~~~~~~~~~~~\texttt{de k+1 ori}\\P(k+1):\underbrace{x\circ x\circ x\circ\ldots\circ x}\circ x=3^{k}(x+1)^{k+1}-1\\~~~~~~~~~~~~~~~~~~~~\texttt{de k ori}\\P(k+1):{[3^{k-1}(x+1)^k-1]}\circ x=3^{k}(x+1)^{k+1}-1\\3{[3^{k-1}(x+1)^k-1+1]}(x+1)-1=3^k(x+1)^{k+1}-1[/tex]
[tex]3\cdot 3^{k-1}(x+1)^k\cdot(x+1)-1=3^{k}\cdot(x+1)^{k+1}-1\\3^{k}(x+1)^{k+1}-1=3^{k}\cdot(x+1)^{k+1}-1,\texttt{adevarat.}\\\texttt{Prin urmare P(k) este adevarat si implicit si P(n) }\\\texttt{este adevarat },\forall n\in\mathbb{N},n\geq 2[/tex]
