I'm not sure if it would change the problem, but one way would make solving it different for me.

I'm not sure if it would change the problem, but one way would make solving it different for me.

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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. – Hero P. · 4 years, 4 months ago

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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. – Mustafa Özçiçek · 4 years, 4 months ago

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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. – Calvin Lin Staff · 4 years, 4 months ago

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– Maddy B · 4 years, 4 months ago

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