anonymous
  • anonymous
Show that (n^2) + 4 < (n + 1)^2 for all natural numbers n>=2
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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schrodinger
  • schrodinger
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anonymous
  • anonymous
for \(n \geq 2\) 4 < 2n +1, right?
anonymous
  • anonymous
I plugged in 2 for n to show the base case.
anonymous
  • anonymous
base 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

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anonymous
  • anonymous
For example, if n =3 , 2n +1 = 7 >4 n =4, 2n +1 = 9 >4 and so on. We all have 2n +1 >4
anonymous
  • anonymous
This should be a mathematical induction proof.
anonymous
  • anonymous
no need
anonymous
  • anonymous
I mean it is required for my homework.
anonymous
  • anonymous
oh!! no need to use induction for this because it is so simple to get the proof
anonymous
  • anonymous
ok, 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
  • anonymous
Now, 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
  • anonymous
Suppose it holds for n = k , that is k^2 + 4 < (k+1)^2
anonymous
  • anonymous
Now prove it holds for n = k+1 that is (k+1)^2 +4 < ((k+1)+1)^2
anonymous
  • anonymous
Yeah, this is where I got stuck.
jim_thompson5910
  • jim_thompson5910
It 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
  • anonymous
ok, 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
  • anonymous
hence \((k+1)^2 +4< k^2 +4k +2 < k^2 +4k +4\) pick far left and far right, we have what we need to prove. ok?
anonymous
  • anonymous
sorry for above, greater than our RIGHT hand side
anonymous
  • anonymous
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anonymous
  • anonymous
Thanks
anonymous
  • anonymous
I understood it.
anonymous
  • anonymous
But, how did u know to add 2k + 1 to both sides?
anonymous
  • anonymous
trick: 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
  • jim_thompson5910
Here'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
  • anonymous
Thanks, to both of you.

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