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so have you tried a proof by induction?

How should I do that with this one?

First step is to show x(1)

Give some more hints please? Cause im just thinking recursive right now, and I dont know what xn is?

this is how your sequence x(n) is defined:
\[x_n=\sqrt{2+x_{n-1}}\] well where x(1)=sqrt(2)

\[x_{n}>0 \text{ for integer } n>0 \\ \\ \text{ so showing } \\ x_nx_{n+1}^2\]

hint write out what x^2(n+2) is

are you still there?

yes, im just trying to understand writing it down

\[x^2_{n+2}=2+x_{n+1}\]
you do understand this is what I'm looking for?

and that we made the assumption that x(n+1)>xn earlier

\[x^2_{n+2}=2+x_{n+1}>2+x_n\]
but guess what 2+x(n) is equal to?

\[x^2_{n+2}=2+x_{n+2}\]

yes plus one...

sorry

\[\sqrt{2}\]?

Yea I cant see the pattern...

\[\text{ we assumed } x_n2+x_n=x_{n+1}^2\]

the end

what do you mean with the end? do you have to the the induction steps with that?

have to do*

but the part you have to do isn't too hard it is just pluggin in n=1 into
\[x_{n+1}=\sqrt{2+x_n}\]

then concluding x(2)>x(1) from that

I already did the other part of the proof

if you don't understand something I did let me know

Yea I will try some on paper now, I will tag you here with questions :) thanks for now!

right which we have done above

we assumed \[x_nx^2_{n+1}\]
remember we did this:
\[x_{n+2}^2=2+x_{n+1}>2+x_n=x_{n+1}^2\]

which you know from that we have x(n+2)>x(n+1) since x^2(n+2)>x^2(n+1)

like you do know that:
\[x^2_{n+2}=2+x_{n+1} \text{ and that } x_{n+1}^2=2+x_n \text{ right ? }\]

\[\text{ if } x_{n+1}>x_n \\ \text{ then } 2+x_{n+1}>2+x_n\]

and if it's any consolation, these things give me a sore head too ;-)

@IrishBoy123, so thats it with no calculation whatsoever? Is that just true..?

Yes

so it isn't just true
we proved it is true by induction

so it is true
I just didn't want you to think we weren't doing anything to prove it
because we did

are you supposed to be learning induction or would any proof do?

Im learning induction!!@IrishBoy123

well, i hope you enjoyed this induction into induction :-)

haha yes, I dont know what I was thinking at first..im gonna close this one now, thanks again guys!