- #1
Chirag B
- 18
- 0
I'm having a little bit of trouble figuring out how exactly to do this.
Prove that [itex]\sqrt{n+\sqrt{n+\sqrt{n+\sqrt{n+\sqrt{n+\cdots}}}}} = \frac{1\pm\sqrt{4n+1}}{2}[/itex].
How exactly does one go about doing this? I mean, I understand it goes on infinitely, but doesn't that create an infinitely large number? Why would the number be so precise?
Also, it appears to be the solution of a quadratic equation (with the [itex]\pm[/itex] and all). Is that an approach? If so, how does one derive such a quadratic equation?
I tried working backwards, but when I do, all I get is that [itex]\sqrt{n+\sqrt{n+\sqrt{n+\sqrt{n+\cdots}}}} = \frac{1\pm\sqrt{4n+1}}{2}[/itex].
There's one less [itex]n[/itex] in the square root, but I still end up getting the same equation. Can this process by repeated infinitely until I get [itex]\sqrt{n} = \frac{1\pm\sqrt{4n+1}}{2}[/itex]? But this isn't true! Can someone help me?
Any help would be gladly appreciated.
Prove that [itex]\sqrt{n+\sqrt{n+\sqrt{n+\sqrt{n+\sqrt{n+\cdots}}}}} = \frac{1\pm\sqrt{4n+1}}{2}[/itex].
How exactly does one go about doing this? I mean, I understand it goes on infinitely, but doesn't that create an infinitely large number? Why would the number be so precise?
Also, it appears to be the solution of a quadratic equation (with the [itex]\pm[/itex] and all). Is that an approach? If so, how does one derive such a quadratic equation?
I tried working backwards, but when I do, all I get is that [itex]\sqrt{n+\sqrt{n+\sqrt{n+\sqrt{n+\cdots}}}} = \frac{1\pm\sqrt{4n+1}}{2}[/itex].
There's one less [itex]n[/itex] in the square root, but I still end up getting the same equation. Can this process by repeated infinitely until I get [itex]\sqrt{n} = \frac{1\pm\sqrt{4n+1}}{2}[/itex]? But this isn't true! Can someone help me?
Any help would be gladly appreciated.