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If $a_n = x_n$ and $a_n = y_n$ are two solutions to the recurrence relation $c_0a_n + c_1a_{n-1}+c_2a_{n-2}+...+c_ka_{n-k} = 0$, show that $z_n = \alpha x_n + \beta y_n$ is also a solution to the recurrence relation for every $\alpha, \beta > 0$. Will the result be true if $\alpha<0$ or $\beta<0$?

Not quite sure, how to approach this problem.

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    $\begingroup$ In the future you will learn the words "linear" and "homogeneous" which make this clear. $\endgroup$ – GEdgar Apr 8 '17 at 18:03
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If I'm not mistaken understanding your question, it suffices to see that the succession $(z_n)_n$ satisfies the given condition. $$ c_0z_n + c_1z_{n-1} + \dots + c_kz_{n-k} = c_0(\alpha x_n + \beta y_n) + c_1(\alpha x_{n-1} + \beta y_{n-1}) + \dots + c_k(\alpha x_{n-k} + \beta y_{n-k}) = \alpha(c_0x_n + c_1x_{n-1}+...+c_kx_{n-k}) + \beta(c_0y_n + c_1y_{n-1}+...+c_ky_{n-k}) = \alpha·0 + \beta ·0 = 0\qquad\forall\alpha,\beta\in\mathbb{R} $$

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    $\begingroup$ Thanks had to review some notes to understand the solution $\endgroup$ – Diante Apr 10 '17 at 3:54

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