Pergunta
a) let t_(1)=2,t_(2)=4,t_(3)=8 and forgeqslant 1,t_(n+3)=t_(n+2)+t_(n+1)+2t_(n) Find a pattern for t_(n) and prove your answer. (4 marks) b) Prove that for any integer x, x is odd if and only if x^2 is odd. (5 marks) c) Prove by contradiction that the sum of a rational number and an irrational number in irrational (6 marks) d) Prove that sqrt (2)+1 is an irrational number (5 marks)
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GustavoProfissional · Tutor por 6 anos
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a) To find a pattern for $t_n$, we can start by calculating the first few terms of the sequence:<br /><br />$t_1 = 2$<br />$t_2 = 4$<br />$t_3 = 8$<br /><br />Now, let's use the given recursive formula to calculate the next terms:<br /><br />$t_4 = t_3 + t_2 + 2t_1 = 8 + 4 + 2(2) = 18$<br />$t_5 = t_4 + t_3 + 2t_2 = 18 + 8 + 2(4) = 38$<br />$t_6 = t_5 + t_4 + 2t_3 = 38 + 18 + 2(8) = 76$<br /><br />From these calculations, we can observe that $t_n = 2^{n-1} + 2^{n-2} + 2^{n-3} + \ldots + 2^2 + 2^1 + 2^0$.<br /><br />To prove this pattern, we can use mathematical induction.<br /><br />Base case: For $n = 1$, we have $t_1 = 2 = 2^{1-1} + 2^{1-2} + 2^{1-3} + \ldots + 2^2 + 2^1 + 2^0$, which is true.<br /><br />Inductive step: Assume that the pattern holds for some $k \geq 1$. We need to show that it holds for $k+1$.<br /><br />Using the recursive formula, we have:<br /><br />$t_{k+1} = t_k + t_{k-1} + 2t_{k-2}$<br /><br />Substituting the pattern for $t_k$, $t_{k-1}$, and $t_{k-2}$, we get:<br /><br />$t_{k+1} = (2^{k-1} + 2^{k-2} + \ldots + 2^2 + 2^1 + 2^0) + (2^{k-2} + 2^{k-3} + \ldots + 2^2 + 2^1 + 2^0) + 2(2^{k-3} + 2^{k-4} + \ldots + 2^2 + 2^1 + 2^0)$<br /><br />Simplifying the expression, we have:<br /><br />$t_{k+1} = 2^{k-1} + 2^{k-2} + \ldots + 2^2 + 2^1 + 2^0 + 2^{k-2} + 2^{k-3} + \ldots + 2^2 + 2^1 + 2^0 + 2^{k-2} + 2^{k-3} + \ldots + 2^2 + 2^1 + 2^0$<br /><br />Combining like terms, we get:<br /><br />$t_{k+1} = 2^{k-1} + 2^{k-2} + \ldots + 2^2 + 2^1 + 2^0$<br /><br />Therefore, the pattern holds for $k+1$, and by the principle of mathematical induction, the pattern holds for all $n \geq 1$.<br /><br />b) To prove that $x$ is odd if and only if $x^2$ is odd, we can use a direct proof.<br /><br />Suppose $x$ is odd. Then, we can write $x = 2k + 1$ for some integer $k$. Squaring both sides, we get:<br /><br />$x^2 = (2k + 1)^2 = 4k^2 + 4k + 1 = 2(2k^2 + 2k) + 1$<br /><br />Since $2k^2 + 2k$ is an integer, $x^2$ is odd.<br /><br />Now, suppose $x^2$ is odd. Then, we can write $x^2 = 2k + 1$ for some integer $k$. Taking the square root of both sides, we get:<br /><br />$x = \sqrt{2k + 1}$<br /><br />Since $2k + 1$ is odd, $x$ must be odd.<br /><br />Therefore, $x$ is odd if and only if $x^2$ is odd.<br /><br />c) To prove by contradiction that the sum of a rational number and an irrational number is irrational, we can assume the opposite, i.e., that the sum of a rational number and an irrational number is rational.<br /><br />Let $r$ be a rational number and $i$ be an irrational number. Suppose their sum $r + i$ is rational. Then, we can write $r + i = q$ for some rational number $
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