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October 19th, 2017, 08:33 AM  #1 
Member Joined: Sep 2014 From: Sweden Posts: 71 Thanks: 0  Induction: Prove that 2^n > (n+2)^2 when n > 6
Prove that $\displaystyle \displaystyle 2^n > (n+2)^2 \ when \ n > 6$ Base case: $\displaystyle \displaystyle P(7): 2^7 > (7+2)^2 = 128 > 81 \ true \\\\ $ Induction Hypothesis:\\ $\displaystyle \displaystyle P(n) \Rightarrow P(n+1)\\\\ \displaystyle Knows: \ 2^n > (n+2)^2\\\\ \displaystyle Wants: \ 2^{n+1}>(n+1+2)^2 = 2^{n+1}>(n+3)^2\\\\ $ Add 5 + 2n to both sides $\displaystyle \displaystyle 5+2n+2^n > n^2+4n+4+5+2n\\\\ \displaystyle 5+2n+2^n > n^2+6n+9\\\\ \displaystyle 5+2n+2^n>(n+3)^2\\\\ \displaystyle 5+2n+(n+2)^2>(n+3)^2\\\\ \displaystyle 5+2n+n^2+4n+4>(n+3)^2\\\\ \displaystyle n^2+6n+9>(n+3)^2\\\\ \displaystyle (n+3)^2>(n+3)^2\\\\ $ This would have worked fine if it was an equal sign $\displaystyle (\displaystyle = or\leq or\geq) $, but it is not. We all know that the same number can't be greater than the same number, it's equal. Is there any great videos or tutorials that explain mathematical induction basics really good? Or an website with induction assignments and so you can see the correct answer/answers? Last edited by DecoratorFawn82; October 19th, 2017 at 08:37 AM. 
October 19th, 2017, 09:43 AM  #2 
Global Moderator Joined: Dec 2006 Posts: 18,155 Thanks: 1422 
Instead of adding $5 + 2n$ to each side, double each side: $2^{n+1} = 2(2^n) > 2(n + 2)^2 = (n + 3)^2 + (n + 3)(n  1) + 2 > (n + 3)^2$ 
October 19th, 2017, 10:27 AM  #3 
Math Team Joined: May 2013 From: The Astral plane Posts: 1,659 Thanks: 652 Math Focus: Wibbly wobbly timeywimey stuff.  
October 19th, 2017, 11:16 AM  #4 
Member Joined: Sep 2014 From: Sweden Posts: 71 Thanks: 0 
I still don't see how we know that $\displaystyle \displaystyle (n+3)^2+(n+3)(n1)+2>(n+3)^2$ though? $\displaystyle \displaystyle (n+3)(n1)+2>0$ That's why right? Last edited by DecoratorFawn82; October 19th, 2017 at 11:33 AM. 
October 19th, 2017, 03:57 PM  #5 
Global Moderator Joined: Dec 2006 Posts: 18,155 Thanks: 1422 
Yes, that's why.


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