anonymous
  • anonymous
prove induction n^2 < n! , for n>=4 , if it is easier n! > n^2 i didnt get far, the basis case is true and (k+1)!= (k+1)k! > ...
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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jamiebookeater
  • jamiebookeater
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watchmath
  • watchmath
I am sorry did you just delete your earlier question with my answer in it?
anonymous
  • anonymous
oh im sorry
anonymous
  • anonymous
this program is strange , i didnt see anything in it

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anonymous
  • anonymous
i apologize earlier, anything you have to say is helpful :)
anonymous
  • anonymous
it doesnt have to be induction
anonymous
  • anonymous
but i have a proof of my own, the basis case is true 4! > 4^2 (k+1)! = k!(k+1) > k^2 ( 2 + 1) = k^2( 1 + 1 + 1) = k^2 + k^2 + k^2 > k^2 + 2k + 1
anonymous
  • anonymous
oh wait, that doesnt work
anonymous
  • anonymous
not the most elegant proof, but it might work
watchmath
  • watchmath
Ok, if you want we can use induction here we go. So we assume that k^2 < k! Notice that that k>1 is equivalent to 2k+1 < 3k It follows that k>= 4 we have 2k+1 < 3k < (k)! k Now using the assumption k^2 < k! we have k^2 +(2k+1) < k! +(2k+1) + k! +(k)! k=(k+1)! So (k+1)^2 < (k+1)!
anonymous
  • anonymous
that looks, good,. one sec, what was your other proof . now im curious
watchmath
  • watchmath
So you haven't read it?
anonymous
  • anonymous
im reading it , one sec
watchmath
  • watchmath
The statement \(n^2n\) The inequality is equivalent to \(n^2-3n+2>n\) \(n^2-4n+2>0\) \(n(n-4)+2>0\) which is clearly true for \(n\geq 4\).
anonymous
  • anonymous
ok thank you
watchmath
  • watchmath
Are you trying to use this for some statement about the series? if it is the case maybe it will be easier to prove the statement for some bigger n (not 4) to make it more obvious.
anonymous
  • anonymous
thats clever, can you look over my proof the basis case is true 4! > 4^2 (k+1)! = k!(k+1) > k^2 ( 2 + 1) = k^2( 1 + 1 + 1) = k^2 + k^2 + k^2 > k^2 + 2k + 1
anonymous
  • anonymous
not that i know of about serious
anonymous
  • anonymous
because k>2 for k>=4
anonymous
  • anonymous
oh but then i need to show k^2 > 2k ?
watchmath
  • watchmath
correct, that is what I was going to say :)
watchmath
  • watchmath
well actually you need 2k^2 > 2k+1
watchmath
  • watchmath
Notice that \((k-1)^2 \geq 0\) So \(k^2-2k+1 \geq 0\) So \(k^2\geq 2k-1\) For \(k\geq 4, k^2 > 2\) So \(2k^2> k^2+2\geq (2k-1)+2 =2k +1\)
anonymous
  • anonymous
yes
anonymous
  • anonymous
can i ask a quick question. im helping a student who thinks that you can form a solid of revolution , bounded by y=x^2, y = x+2, and revolve it about the y axis
watchmath
  • watchmath
Yes I think you can but there is some overlap between the solid form by the region on the right of the y-axis and the solid form by the region on the left side of the y-axis
anonymous
  • anonymous
ty
anonymous
  • anonymous
keeps freezing on me

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