Melody

Because there is a very real community of low level nerds here (I am including myself), we enjoy each others company, we like the stimulation of the questions, we like helping people to learn and we have fun. 

Sorry anon I think you screwed it up - just a little

-6(1-2r)+7

=-6 - -12r+7

=-6 + 12r+7

=-6+7+12r

=1+12r

Thanks Chris and Brittany  

My pleasure Chris :)

Well Bertie I have given it a shot   

$$\\\mbox{Prove }\sqrt{1+x\sqrt{1+(x+1)\sqrt{1+(x+2)\sqrt{.....}}}}=x+1\\
\\LHS\\
f(x)=\sqrt{1+x\sqrt{1+(x+1)\sqrt{1+(x+2)\sqrt{.....}}}}\\
f(x+1)=\sqrt{1+(x+1)\sqrt{1+(x+2)\sqrt{.....}}}\\\\
f(x)=\sqrt{1+x*f(x+1)}\\\\
f^2(x)=1+x[f(x+1)]$$

$$\\RHS\\
g(x)=x+1\\
g(x+1)=x+1+1\\
g(x+1)=1+g(x)\\
g^2(x)=(x+1)^2\\
g^2(x)=x^2+2x+1\\
g^2(x)=1+x(x+2)\\
g^2(x)=1+x[g(x+1)]\\$$

$$\\f^2(x)=1+x[f(x+1)]\qquad and\qquad g^2(x)=1+x[g(x+1)]\\\\
$It follows that$\\\\
f^2(x)=g^2(x)\\\\
Now f(x)\ge 0 \qquad $because it is the answer to a square root$\\\\
g(x)=x+1 $ which \;must\; be\; $\ge 0 $ Therefore x is greater than or equal to -1$\\\\$$

$$\mbox{The statement cannot be true if x is less than -1 because the}\\ \mbox{RHS would be neg and the Left hand side can never be negative}$$

$$\\$so for all real values of x greater than or equal to -1$\\
\\f(x)=g(x) \\\\
\sqrt{1+x\sqrt{1+(x+1)\sqrt{1+(x+2)\sqrt{.....}}}}=x+1\qquad Q.E.D.\\$$

As an aside:

$$\\\mbox{A look at two individual cases }\\
If\;\; x=0 \;then\\
LHS=\sqrt{1+0*...}=\sqrt1=1=RHS\\\\
If\;\; x=-1 \;then\\
LHS=\sqrt{1+-1*\sqrt{1+(-1+1)\sqrt...}}=
\sqrt{1+-1*\sqrt{1+0}}=\sqrt{1+-1*1}=\sqrt{0}=0=RHS\\\\$$

I am really not completely sure of of this proof.

Is it really true for ALL real values greater of equal to -1?

Is it completely correct Bertie or does it need refining?

  @@ End of Day Wrap          Fri 6/2/15 Sydney,        Australia Time      11:20 pm          ♪ ♫    

Hi Everyone,

Zegroes, stay out of my wrap post ! If you are a major pest you will be treated like your poor friend Louie.

Seriously now - I do not mind people adding the occasional relevant post here but it is not a play thread.

Today our wonderful answers were supplied by Heureka, CPhill, Trincent, Happy7, geno3141, Alan, Takahiromaeda, BrittanyJ, Mahun, GenericUserName, Zegroes and saseflower.  My thanks to each of you. 

Now I think I collected a number of interest posts but many appear to be lost.  Perhaps I closed without saving.         Anyway, here are a couple.

Sat 7/2/15

You mistake is that

$$(x-3)^2=x^2-6x+9\\\\
$It does not equal $-x^2+6x-9\\\
$that is why -1 had to be left out of the bracket!$$$

$$g(x+2)=\sqrt{x+2}$$