Solving the Identity Challenge: $3=\sqrt{1+2...9}$

[Albert1]

prove:
$3=\sqrt{1+2\sqrt{1+3\sqrt{1+4\sqrt{1+5\sqrt{1+6\sqrt{1+7\sqrt{1+8\sqrt{1+9--}}}}}}}}$

 

[Albert1]

prove:
$3=\sqrt{1+2\sqrt{1+3\sqrt{1+4\sqrt{1+5\sqrt{1+6\sqrt{1+7\sqrt{1+8\sqrt{1+9--}}}}}}}}$

Spoiler

let:$f(x)=x(x+2)---(1)$
then: $f(x+1)=(x+1)(x+3)=(x+2)^2-1$
$\therefore (x+2)=\sqrt {1+f(x+1)}---(2)$
from (1)and (2):
$f(x)=x(x+2)=x\sqrt{1+f(x+1)}---(3)$
using (3) recursively we get :
$f(x)=x\sqrt{1+(x+1)\sqrt{1+f(x+2)}}$
$=x\sqrt{1+(x+1)\sqrt{1+(x+2)\sqrt{1+f(x+3)---}}}=x(x+2)$
let:$x=1$
and we get the result

 

[topsquark]

I have to take a moment and "Oooh!" and "Ahhhh!" over this.

Oooh!

Ahhhh!

Nice one! (Bow)

-Dan