My Math Forum Induction: Prove that 2^n > (n+2)^2 when n > 6
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 October 19th, 2017, 07:33 AM #1 Member   Joined: Sep 2014 From: Sweden Posts: 92 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 07:37 AM.
 October 19th, 2017, 08:43 AM #2 Global Moderator   Joined: Dec 2006 Posts: 19,542 Thanks: 1752 Instead of adding $5 + 2n$ to each side, double each side: $2^{n+1} = 2(2^n) > 2(n + 2)^2 = (n + 3)^2 + (n + 3)(n - 1) + 2 > (n + 3)^2$ Thanks from topsquark
October 19th, 2017, 09:27 AM   #3
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Quote:
 Originally Posted by DecoratorFawn82 $\displaystyle \displaystyle P(7): 2^7 > (7+2)^2 = 128 > 81 \ true \\\\$
A quick notation comment. The equal sign in the above is confusing. I'd recommend $\displaystyle \rightarrow$ or $\displaystyle \Rightarrow$ instead of the =.

-Dan

 October 19th, 2017, 10:16 AM #4 Member   Joined: Sep 2014 From: Sweden Posts: 92 Thanks: 0 I still don't see how we know that $\displaystyle \displaystyle (n+3)^2+(n+3)(n-1)+2>(n+3)^2$ though? $\displaystyle \displaystyle (n+3)(n-1)+2>0$ That's why right? Last edited by DecoratorFawn82; October 19th, 2017 at 10:33 AM.
 October 19th, 2017, 02:57 PM #5 Global Moderator   Joined: Dec 2006 Posts: 19,542 Thanks: 1752 Yes, that's why.

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