<s>[INST] Find $f_{10}$ of the sequence defined by $f_{n+1}=\frac{8f_n}{5}+\frac{6 \sqrt{4^n-f_n^2}}{5}$ Find $f_{10}$ of the sequence defined by $$f_{n+1}=\frac{8f_n}{5}+\frac{6 \sqrt{4^n-f_n^2}}{5}$$ given $f_0=0$
My Approach:
Letting $f_n=2^n b_n$ we get
$$b_{n+1}=\frac{4b_n}{5}+\frac{3 \sqrt{1-b_n^2}}{5}$$
Now letting $b_n=\cos(x_n)$ we get
$$\cos(x_{n+1})=\cos(x_n-\theta)$$ where $\cos(\theta)=\frac{4}{5}$
Now Since $f_0=0$ we have $b_0=0$ and $x_0=\frac{\pi}{2}$
Now we have
$$x_{n+1}=x_n-\theta$$
Putting $n=0,1,2,3 \cdots 10$ and adding all we get
$$x_{10}=\frac{\pi}{2}-10\theta$$
Hence
$$b_{10}=\cos\left(\frac{\pi}{2}-10\theta\right)=\sin(10\theta)=\sin\left(10\arcsin\left(\frac{3}{5}\right)\right)$$
How to proceed further?
 [/INST] Simply writing it out by hand quickly yields
$$f_1=\frac{6}{5},\qquad f_2=\frac{96}{25},\qquad f_3=\frac{936}{125},\qquad f_4=\frac{384}{25},$$
and the rest should only take a few minutes. Here's a few more, it's not that hard:
$$f_5=\frac{3744}{125},\qquad f_6=\frac{1536}{25},\qquad f_7=\frac{14976}{125},\qquad f_8=\frac{6144}{25}.$$
I must say that $f_8$ and $f_{10}$ took slightly longer, but all in all this is no more than ten minutes work:
$$f_9=\frac{59904}{125},\qquad f_{10}=\frac{24576}{25}.$$
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