let $a_{1}=\dfrac{\sqrt{2}}{4}$ and such $$a_{n+1}=\sqrt{2a_{n}+1}$$ find $a_{n}$
my idea:let $a_{n}=\dfrac{1}{2}\cos{x_{n}}$ $$\Longrightarrow \dfrac{1}{2}\cos{x_{n+1}}=\sqrt{\cos{x_{n}}+1}=\sqrt{2}\cos{\dfrac{x_{n}}{2}}$$
following I can't work.
But I have see this $$a_{1}=\sqrt{2},a_{n+1}=\sqrt{2+a_{n}}$$ This can let $a_{n}=2\cos{x_{n}}$ can work
bceause $$\Longrightarrow 2\cos{x_{n+1}}=\sqrt{2+2\cos{x_{n}}}=2\cos{\dfrac{x_{n}}{2}}$$ so $$\cos{x_{n+1}}=\cos{x_{n}/2},x_{1}=\dfrac{\pi}{4}$$
so $$x_{n}=2\cos{\dfrac{\pi}{2^{n-1}}}$$
Note that if $a_n \gt 0$ then $a_{n+1}\gt 1$. Also if $a_n \lt 3$ then $a_{n+1} \lt 3$. So that bounds the sequence.
The fixed point of the sequence satisfies $a=\sqrt {2a+1}$ - square and solve to obtain $a=1+\sqrt 2$.
It is easy to prove that $a_n$ is increasing to this fixed point.
If you are hoping to guess a general expression for $a_n$ then it has to meet these constraints. However even calculating $a_2$ by hand things look to be getting messy - note: when stuck do work out the first few values.