If one solves the recurrence relation \( a_{n+2} = k_1 a_{n+1} + k_0 a_n \) for suitable constants \( k_0, k_1 \) such that it holds for all positive integers \( n \), then it also holds for all real numbers \( n \). Therefore, it makes no difference whether one considers the case \( n = 0 \) in the solution; either way, the same constants will be obtained.

To see why this is the case, we note that if \( \sin \theta = \frac{7}{25} \), with \( 0 < \theta < \frac{\pi}{2} \), then \( \cos \theta = \frac{24}{25} \). Then \[ \begin{align*} \sin 2\theta &= 2 \sin \theta \cos \theta = \frac{336}{625}, \\ \cos 2\theta &= 1 - 2\sin^2 \theta = \frac{527}{625}. \end{align*} \] Hence \[ \begin{align*} a_{n+2} &= \sin n \theta \cos 2\theta + \cos n \theta \sin 2\theta \\ &= \frac{527}{625} \sin n\theta + \frac{336}{625} \cos n\theta. \end{align*} \] But we also have \[ \begin{align*} a_{n+2} &= k_1 a_{n+1} + k_0 a_n \\ &= k_1 (\sin n \theta \cos \theta + \cos n \theta \sin \theta) + k_0 \sin n \theta \\ &= \left( k_0 + \frac{24k_1}{25} \right) \sin n\theta + \frac{7k_1}{25} \cos n\theta. \end{align*} \] So if this recurrence is to be satisfied for positive integers \( n \), then it is immediately obvious that it is also satisfied for arbitrary real \( n \), since the equality of the coefficients gives identical equality. In particular, we must have \[ \begin{align*} k_0 + \frac{24}{25} k_1 &= \frac{527}{625}, \\ \frac{7}{25} k_1 &= \frac{336}{625}, \end{align*} \] the solution of which is \( k_0 = -1 \), \( k_1 = \frac{48}{25} \). Therefore, \( k_0 + k_1 = \frac{23}{25} \). However, we should note that if we were to approach the question by substituting in \( n = 0 \), the resulting equation is underdetermined for \( k_0 \). That doesn't mean the recurrence is false for this value; it simply means that \( k_0 \) is not uniquely determined in that specific case.

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TopNewestNow that the question is closed, I can respond.

If one solves the recurrence relation \( a_{n+2} = k_1 a_{n+1} + k_0 a_n \) for suitable constants \( k_0, k_1 \) such that it holds for all positive integers \( n \), then it also holds for all real numbers \( n \). Therefore, it makes no difference whether one considers the case \( n = 0 \) in the solution; either way, the same constants will be obtained.

To see why this is the case, we note that if \( \sin \theta = \frac{7}{25} \), with \( 0 < \theta < \frac{\pi}{2} \), then \( \cos \theta = \frac{24}{25} \). Then \[ \begin{align*} \sin 2\theta &= 2 \sin \theta \cos \theta = \frac{336}{625}, \\ \cos 2\theta &= 1 - 2\sin^2 \theta = \frac{527}{625}. \end{align*} \] Hence \[ \begin{align*} a_{n+2} &= \sin n \theta \cos 2\theta + \cos n \theta \sin 2\theta \\ &= \frac{527}{625} \sin n\theta + \frac{336}{625} \cos n\theta. \end{align*} \] But we also have \[ \begin{align*} a_{n+2} &= k_1 a_{n+1} + k_0 a_n \\ &= k_1 (\sin n \theta \cos \theta + \cos n \theta \sin \theta) + k_0 \sin n \theta \\ &= \left( k_0 + \frac{24k_1}{25} \right) \sin n\theta + \frac{7k_1}{25} \cos n\theta. \end{align*} \] So if this recurrence is to be satisfied for positive integers \( n \), then it is immediately obvious that it is also satisfied for arbitrary real \( n \), since the equality of the coefficients gives identical equality. In particular, we must have \[ \begin{align*} k_0 + \frac{24}{25} k_1 &= \frac{527}{625}, \\ \frac{7}{25} k_1 &= \frac{336}{625}, \end{align*} \] the solution of which is \( k_0 = -1 \), \( k_1 = \frac{48}{25} \). Therefore, \( k_0 + k_1 = \frac{23}{25} \). However, we should note that if we were to approach the question by substituting in \( n = 0 \), the resulting equation is underdetermined for \( k_0 \). That doesn't mean the recurrence is false for this value; it simply means that \( k_0 \) is not uniquely determined in that specific case.

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It does not, I couldn't solve it starting with n+2=2.

Try it without including it. It should've said it in the description, though. It took a lot of time for me to see that.

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Note that your 'conclusion' is false.

You merely tried to find values which satisfy 1 of the many equations, and did not find a set of values which satisfied ALL of the equations.

Read the solution, which is similar to what Hero posted above.

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Thanks

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