Mathematical Induction
A sequence is defined by un+1=(2+un)^(1/2) and u1=3.
Prove by mathematical induction un>2 for all positive integers n.
Prove by mathematical induction un>2 for all positive integers n.
Original post by Arnoldedward
A sequence is defined by un+1=(2+un)^(1/2) and u1=3.
Prove by mathematical induction un>2 for all positive integers n.
Prove by mathematical induction un>2 for all positive integers n.
What steps have you done so far? Can you show us your working?
Let P(n) be the proposition that un>2.
When n=1, u1=3 >2
P(1) is true.
Assume P(k) is true, then uk>2.
uk+1=(2+uk)^(1/2)
After that, I have no idea how to continue.
When n=1, u1=3 >2
P(1) is true.
Assume P(k) is true, then uk>2.
uk+1=(2+uk)^(1/2)
After that, I have no idea how to continue.
Original post by Arnoldedward
Assume P(k) is true, then uk>2.
uk+1=(2+uk)^(1/2)
Assume P(k) is true, then uk>2.
uk+1=(2+uk)^(1/2)
If uk is >2, then what must the RHS of your inequality be greater than?
Original post by ghostwalker
If uk is >2, then what must the RHS of your inequality be greater than?
If uk>2, the RHS will be greater than 2.
but how to show the workings??
Original post by Arnoldedward
If uk>2, the RHS will be greater than 2.
but how to show the workings??
but how to show the workings??
IF uk >2
then uk+2 > 4
Hence, sqrt(uk+2) > 2
I got it.
Thank you guys.
Thank you guys.
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