GRÊMIO X CORITIBA: A BATALHA TRICOLOR RUMO À GLÓRIA!
GRÊMIO E CORITIBA: UM CLÁSSICO COM HISTÓRIA
Grêmio e Coritiba se enfrentam em um duelo com muita história. Ao longo dos anos, as duas equipes protagonizaram grandes jogos e disputas memoráveis. O confronto entre gaúchos e paranaenses é sempre um evento aguardado pelos torcedores, que esperam um espetáculo de futebol.
GRÊMIO: EM BUSCA DA REABILITAÇÃO
O Grêmio entra em campo pressionado paraFind a sequence $a_n$ such that $a_0=1, a1=0, a{n+2}=-a_{n+1}+an$ Let the given recurrence relation be $a{n+2} = -a_{n+1} + a_n$ with initial conditions $a_0=1, a_1=0$. We can find the first few terms of the sequence: $a_0 = 1$ $a_1 = 0$ $a_2 = -a_1 + a_0 = -0 + 1 = 1$ $a_3 = -a_2 + a_1 = -1 + 0 = -1$ $a_4 = -a_3 + a_2 = -(-1) + 1 = 1+1 = 2$ $a_5 = -a_4 + a_3 = -2 + (-1) = -3$ $a_6 = -a_5 + a_4 = -(-3) + 2 = 3+2 = 5$ $a_7 = -a_6 + a_5 = -5 + (-3) = -8$ The sequence is $1, 0, 1, -1, 2, -3, 5, -8, \dots$ Let’s assume the solution is of the form $a_n = r^n$ for some $r$. Substituting this into the recurrence relation, we get $r^{n+2} = -r^{n+1} + r^n$ Dividing by $r^n$, we get the characteristic equation $r^2 = -r + 1$ $r^2 + r – 1 = 0$ The roots are $r = \frac{-1 \pm \sqrt{1^2 – 4(1)(-1)}}{2(1)} = \frac{-1 \pm \sqrt{1+4}}{2} = \frac{-1 \pm \sqrt{5}}{2}$. Let $r_1 = \frac{-1+\sqrt{5}}{2}$ and $r_2 = \frac{-1-\sqrt{5}}{2}$. Then the general solution is $a_n = A r_1^n + B r_2^n = A \left( \frac{-1+\sqrt{5}}{2} \right)^n + B \left( \frac{-1-\sqrt{5}}{2} \right)^n$ for some constants $A$ and $B$. We can find $A$ and $B$ using the initial conditions. $a_0 = 1 = A \left( \frac{-1+\sqrt{5}}{2} \right)^0 + B \left( \frac{-1-\sqrt{5}}{2} \right)^0 = A + B$ $a_1 = 0 = A \left( \frac{-1+\sqrt{5}}{2} \right)^1 + B \left( \frac{-1-\sqrt{5}}{2} \right)^1 = A \left( \frac{-1+\sqrt{5}}{2} \right) + B \left( \frac{-1-\sqrt{5}}{2} \right)$ From $A+B=1$, we have $B = 1-A$. Substituting this into the second equation, we get $0 = A \left( \frac{-1+\sqrt{5}}{2} \right) + (1-A) \left( \frac{-1-\sqrt{5}}{2} \right) = A \left( \frac{-1+\sqrt{5}}{2} \right) + \frac{-1-\sqrt{5}}{2} – A \left( \frac{-1-\sqrt{5}}{2} \right)$ $A \left( \frac{-1+\sqrt{5}}{2} + \frac{1+\sqrt{5}}{2} \right) = \frac{1+\sqrt{5}}{2}$ $A \left( \frac{2\sqrt{5}}{2} \right) = \frac{1+\sqrt{5}}{2}$ $A \sqrt{5} = \frac{1+\sqrt{5}}{2}$ $A = \frac{1+\sqrt{5}}{2\sqrt{5}} = \frac{\sqrt{5}+5}{10}$ $B = 1-A = 1 – \frac{1+\sqrt{5}}{2\sqrt{5}} = \frac{2\sqrt{5}-(1+\sqrt{5})}{2\sqrt{5}} = \frac{-\sqrt{5}+1}{2\sqrt{5}} = \frac{-\sqrt{5}+1}{2\sqrt{5}} \frac{\sqrt{5}}{\sqrt{5}} = \frac{-5+\sqrt{5}}{10}$ Thus, $A=\frac{5+\sqrt{5}}{10}$, $B=\frac{1-\sqrt{5}}{2\sqrt{5}} = \frac{\sqrt{5}-5}{10}$ So, $a_n = \frac{5+\sqrt{5}}{10} \left( \frac{-1+\sqrt{5}}{2} \right)^n + \frac{5-\sqrt{5}}{10} \left( \frac{-1-\sqrt{5}}{2} \right)^n$ Let $\phi = \frac{1+\sqrt{5}}{2}$ and $\psi = \frac{1-\sqrt{5}}{2}$ So $r_1 = -\psi, r_2 = -\phi$ $a_n = \frac{5+\sqrt{5}}{10} (-\psi)^n + \frac{5-\sqrt{5}}{10} (-\phi)^n$
Final Answer: The final answer is $\boxed{a_n}$
