Question

%20%7D%20%7D%20%7D%20" id="TexFormula1" title=" \sqrt{6 + \sqrt{6 + \sqrt{6 + ..... \infty = } } } " alt=" \sqrt{6 + \sqrt{6 + \sqrt{6 + ..... \infty = } } } " align="absmiddle" class="latex-formula">
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Answer 1

Answer:

3

Step-by-step explanation:

hello,

let's define the following sequence

$a_0=\sqrt{6}\\a_{n+1}=\sqrt{6+a_n}$

the number we are looking for is the limit of the sequence

we can prove that the sequence is increasing and

$a_n$ by induction

(by noticing that if $a_n$ then  $a_{n+1}=\sqrt{6+a_n}$ )

so the limit of the sequence exists, let's note it l we can write from

$a_{n+1}=\sqrt{6+a_n}$

that

$l=\sqrt{6+l}$

so

$l^2=6+l l^2-l-6=0 l^2-3l+2l-6=0\\ l(l-3)+2(l-3)=0(l+2)(l-3)=0\\\\l=-2 \ \ or \ \ l = 3$

as

$a_n >= 0$ it comes l = 3

hope this helps

Answer 2

Answer:

$3$

Step-by-step explanation:

$\sqrt{6+\sqrt{6+\sqrt{6+\sqrt{6+\sqrt{6+\sqrt{6+\sqrt{6+\sqrt{6+\sqrt{6...}}}}}}}}}$

$= 2.999999999999999999999999999...$

$= 3$