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anonymous
 one year ago
Show that (n^2) + 4 < (n + 1)^2 for all natural numbers n>=2
anonymous
 one year ago
Show that (n^2) + 4 < (n + 1)^2 for all natural numbers n>=2

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anonymous
 one year ago
Best ResponseYou've already chosen the best response.0for \(n \geq 2\) 4 < 2n +1, right?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I plugged in 2 for n to show the base case.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0base case n =2, 4 < 2n +1 =5, check the left hand side is 4, a constant, never change. while the right hand side is increasing with n > 2

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0For example, if n =3 , 2n +1 = 7 >4 n =4, 2n +1 = 9 >4 and so on. We all have 2n +1 >4

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0This should be a mathematical induction proof.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I mean it is required for my homework.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0oh!! no need to use induction for this because it is so simple to get the proof

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0ok, I show you this first 4 < 2n +1 +n^2 both sides you have n^2 +2 < n^2 +2n +1 the right hand side is (n+1)^2 done.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Now, if you want to use induction, ok, go ahead basic case n =2 n^2 +4 = 8 < (n+1)^2 =(2+1)^2 =9 check

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Suppose it holds for n = k , that is k^2 + 4 < (k+1)^2

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Now prove it holds for n = k+1 that is (k+1)^2 +4 < ((k+1)+1)^2

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Yeah, this is where I got stuck.

jim_thompson5910
 one year ago
Best ResponseYou've already chosen the best response.1It might help to think of `k^2 + 4 < (k + 1)^2` as `k^2 + 4 + q = (k + 1)^2` where `q` is some positive number

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0ok, from k^2 +4 < k^2 +2k+1 add 2k +1 both sides k^2 + 2k +1 +4 < k^2 + 2k +1 +2k +1 (k+1)^2 +4 < k^2 +4k +2 If we consider (k+2)^2 = k^2 + 4k + 4 which is greater than our left hand side

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0hence \((k+1)^2 +4< k^2 +4k +2 < k^2 +4k +4\) pick far left and far right, we have what we need to prove. ok?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0sorry for above, greater than our RIGHT hand side

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0dw:1442708748614:dw

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0But, how did u know to add 2k + 1 to both sides?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0trick: open the left hand side of the expression we need to prove, that is (k+1) ^2 +4 = k^2 +2k +1 +4 we can see that we need 2k +1

jim_thompson5910
 one year ago
Best ResponseYou've already chosen the best response.1Here's an alternative method to do the inductive step. I'm attaching it as a txt file so I don't barge in too much and add more clutter to the page.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Thanks, to both of you.
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