General term of $a_{n}= \sqrt{ a_{n-1}+6}$ What is the general term of  $a_{n}= \sqrt{ a_{n-1}+6}$,$a_{1}=4$?
 A: Working out a few elements of the sequence:
$$a_1 = 4 \\
a_2 = \sqrt {10} \\
a_3 = \sqrt {6 + \sqrt {10}} \\
a_4 = \sqrt {6 + \sqrt {6 + \sqrt{10}}} \\
\vdots \\
a_n = \sqrt {6 + \sqrt {6 + \sqrt {\cdots + \sqrt{10} }} }
$$
I am guessing you can't easily get a "closed-form" expression for the general term. (The reason I guess this is that the minimal degree of the field extension of $\Bbb{Q}$ that contains $a_n$ is unbounded with $n$. If someone reading this disagrees with me I would like to hear it :) )
However, maybe you don't need the general term. I remember having exercises that look like this one in calculus courses, where the question wasn't "find the general term" but rather "find the limit". If that's your question, here's a strategy:


*

*Figure out what the possible limits are. (Assume the sequence has a limit and use the recurrence to rule out all values except two.)

*Guess what it really is (i.e. using a calculator).

*Show that the sequence is monotonous (descending).

*Show that the sequence is always greater than the guessed limit. (Use induction.)

*Deduce the limit from the above.

A: I doubt if such general closed form exists. 
We have $\sqrt{a_{n-1}+6}$ to be $\sqrt{10}$ in the first step, and $\sqrt{6+\sqrt{10}}$ is hard to simplify as if one assume it has the form $a+\sqrt{b}$, then we must have $a^{2}+b=6,2a\sqrt{b}=\sqrt{10}$. This force $a^{2}b=5/2$, and one has to solve $x^{2}-6x+5/2$. This gives $a^{2}$ or $b$ equal to $\frac{6+\sqrt{36-10}}{2}=3+\frac{\sqrt{23}}{2}$, which is fairly ugly. Another guess might be $\sqrt{c}+\sqrt{d}$, but then we would have $6=c+d$ and $10=4cd$, which ends up to the same thing. Since $a_{2}$ is already such a mess it is hard to come up with general formula for $a_{i}$. 
