Halo every one, please help me to solve this question.

[AgungSusan]

 

[Romsek]

Well... the answer is 3 but I can't prove it yet.

 

[Zermelo]

View attachment 24260

You need to find a sequence that describes the expression in the picture.
a1 (the first element in the sequence) = [MATH]\sqrt{4}[/MATH]a2 = [MATH]\sqrt{4 + \sqrt{16} }[/MATH]a3 = [MATH]\sqrt{4 + \sqrt{ 16 + \sqrt {64} } }[/MATH]...
an = ?
You need to find what the "n-th" member of the sequence is. Then, because you have the "...", it means that the sequence goes on without an end (English isn't my first language but you'll understand what I mean), so you need to find [MATH]\lim_{n\to\infty} a_n[/MATH].
If you find a nice expression for a_n it will be easy, but you'll also maybe find that the sequence is defined recursively.
For example: [MATH]\sqrt{2+\sqrt{2+\sqrt{2 + ...} }}[/MATH] can be described with the sequence:
[MATH]a_1 = \sqrt{2}[/MATH][MATH]a_{n+1} = \sqrt{2+a_n}[/MATH]If you find this way of defining the sequence, there will be some more work to be done to prove that this sequence actually converges (Weierstrass theorem), but then you just look at the way the sequence is defined, take the limit of both sides and voila. For exaple:
[MATH]a_{n+1} = \sqrt{2+a_n}[/MATH]It should be clear that the limit of a_n is the same as limit of a_n+1, so when we take the limit of both sides we have:
[MATH]\lim a_{n+1} = \lim \sqrt{2+a_n}[/MATH]Let's denote [MATH]\lim a_n = \lim a_{n+1} = a[/MATH]So we have
[MATH]a = \lim \sqrt{2+a_n}[/MATH][MATH]a = \sqrt{2+a}[/MATH]which is now trivial to solve.
The hard part is figuring out the expression for the sequence.