Question

$a_0 = 0$

$a_{n+1} = \frac{8}{5}a_n + \frac{6}{5}\sqrt{4^n - a_n^2}$

$[a_{10}] = ?$

Answer 1

983

Answer 2

Let $a_n = 2^n \sin(\theta_n)$. Substituting this into the recurrence relation, we get: $2^{n+1} \sin(\theta_{n+1}) = \frac{8}{5} (2^n \sin(\theta_n)) + \frac{6}{5} \sqrt{4^n - (2^n \sin(\theta_n))^2}$ $2^{n+1} \sin(\theta_{n+1}) = \frac{8}{5} 2^n \sin(\theta_n) + \frac{6}{5} 2^n |\cos(\theta_n)|$ Dividing by $2^{n+1}$: $\sin(\theta_{n+1}) = \frac{4}{5} \sin(\theta_n) + \frac{3}{5} |\cos(\theta_n)|$

Let $\alpha$ be an angle such that $\cos(\alpha) = \frac{4}{5}$ and $\sin(\alpha) = \frac{3}{5}$. The relation becomes: $\sin(\theta_{n+1}) = \cos(\alpha) \sin(\theta_n) + \sin(\alpha) |\cos(\theta_n)|$

With $a_0 = 0$, we have $\theta_0 = 0$. The sequence of angles $\theta_n$ and their cosines are: $\theta_0 = 0$, $\cos(\theta_0) = 1 > 0$. $\sin(\theta_1) = \sin(0+\alpha) = \sin(\alpha) \implies \theta_1 = \alpha$. $\cos(\theta_1) = 4/5 > 0$. $\sin(\theta_2) = \sin(\alpha+\alpha) = \sin(2\alpha) \implies \theta_2 = 2\alpha$. $\cos(\theta_2) = \cos(2\alpha) = 7/25 > 0$. $\sin(\theta_3) = \sin(2\alpha+\alpha) = \sin(3\alpha)$. $\cos(3\alpha) = -44/125 < 0$. To ensure the cosine term is positive for the next step if needed, we choose $\theta_3 = \pi - 3\alpha$. $\cos(\theta_3) = \cos(\pi - 3\alpha) = 44/125 > 0$. $\sin(\theta_4) = \sin(\theta_3+\alpha) = \sin(\pi - 3\alpha + \alpha) = \sin(\pi - 2\alpha) = \sin(2\alpha)$. $\cos(\theta_4) = \cos(\pi - 2\alpha) = -7/25 < 0$. $\sin(\theta_5) = \sin(\theta_4-\alpha) = \sin(\pi - 2\alpha - \alpha) = \sin(\pi - 3\alpha) = \sin(3\alpha)$. $\cos(\theta_5) = \cos(\pi - 3\alpha) = 44/125 > 0$.

The pattern for $n \ge 3$ is: If $n$ is odd, $\theta_n = \pi - 3\alpha$. If $n$ is even, $\theta_n = \pi - 2\alpha$.

We need $a_{10} = 2^{10} \sin(\theta_{10})$. Since 10 is even and $10 \ge 4$, $\theta_{10} = \pi - 2\alpha$. $a_{10} = 2^{10} \sin(\pi - 2\alpha) = 2^{10} \sin(2\alpha)$. $\sin(2\alpha) = 2\sin(\alpha)\cos(\alpha) = 2(3/5)(4/5) = 24/25$. $a_{10} = 1024 \times \frac{24}{25} = \frac{24576}{25} = 983.04$. The integer part is $[a_{10}] = [983.04] = 983$.