The sequence {b_{n}} satisfies b_{0}=0 and

\color{brown}\displaystyle 5b_{n+1}=8b_{n}+6\sqrt{4^n-b^2_{n}} for n\geq 0.Then value of \color{brown}\lfloor b_{10}\rfloor is

@Hari_Shankar Bhatt Sir, plz have a look on that problem,

Sir , Thanks

The sequence {b_{n}} satisfies b_{0}=0 and

\color{brown}\displaystyle 5b_{n+1}=8b_{n}+6\sqrt{4^n-b^2_{n}} for n\geq 0.Then value of \color{brown}\lfloor b_{10}\rfloor is

@Hari_Shankar Bhatt Sir, plz have a look on that problem,

Sir , Thanks

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Write as

\dfrac{b_{n+1}}{2^{n+1}} = \left(\dfrac{8}{10} \right) \dfrac{b_n}{2^n}+\dfrac{6}{10} \sqrt{1-\dfrac{b}{2^n}}

Consider a_n = \dfrac{b_n}{2_n}

Then the recurrence can be written as

a_{n+1} = \left(\dfrac{4}{5} \right) a_n+ \left(\dfrac{3}{5} \right) \sqrt{1-a_n^2}

With a_n = \sin \theta,a_0=0 and \sin \alpha = \dfrac{3}{5} the sequence is a_n=\sin (n \alpha)

So b_{10} = 2^{10} \sin (10 \alpha) Using the known estimate of \alpha = 37^{\circ}, i get \lfloor b_{10} \rfloor = 177

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