Suppose that a given sequence $\{x_n\}_{n\geq 1}$ satisfies the conditions that $x_1=1$ and, for $n\geq 1$, $$x_{n+1}=\frac{1}{16}(1+4x_{n}+\sqrt{1+24x_{n}}).$$ Determine $\displaystyle \lim_{n\to+\infty} 3x_n.$

I tried writing out the next few terms but I don't see how to do any cancellation. Any small clues? Thanks.

Hint. If the limit exists and it is finite, say $\displaystyle \lim_{n\to+\infty} x_n=L\in\mathbb{R}$, then by continuity, $$16L=1+4L+\sqrt{1+24L}\implies L=?$$ In order to show the convergence consider the increasing function $$f(x)=\frac{1}{16}(1+4x+\sqrt{1+24x})$$ and show that it is a contraction in $[f(0),+\infty)=[1/8,+\infty)$ by evaluating its derivative.

P.S. The sequence $\{x_n\}_{n\geq 1}$ converges to the unique fixed point $L$ for any $x_1\geq 0$.

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