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I am sorry did you just delete your earlier question with my answer in it?
oh im sorry
this program is strange , i didnt see anything in it
i apologize earlier, anything you have to say is helpful :)
it doesnt have to be induction
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
oh wait, that doesnt work
not the most elegant proof, but it might work
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)!
that looks, good,. one sec, what was your other proof . now im curious
So you haven't read it?
im reading it , one sec
The statement \(n^2
The inequality is equivalent to
which is clearly true for \(n\geq 4\).
ok thank you
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.
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
not that i know of about serious
because k>2 for k>=4
oh but then i need to show k^2 > 2k ?
correct, that is what I was going to say :)
well actually you need 2k^2 > 2k+1
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 \(2k^2> k^2+2\geq (2k-1)+2 =2k +1\)
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
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
keeps freezing on me