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October 19th, 2017, 08:33 AM   #1
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Question 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.
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October 19th, 2017, 09:43 AM   #2
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Instead of adding $5 + 2n$ to each side, double each side:

$2^{n+1} = 2(2^n) > 2(n + 2)^2 = (n + 3)^2 + (n + 3)(n - 1) + 2 > (n + 3)^2$
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October 19th, 2017, 10:27 AM   #3
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Quote:
Originally Posted by DecoratorFawn82 View Post
$\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
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October 19th, 2017, 11:16 AM   #4
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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 11:33 AM.
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October 19th, 2017, 03:57 PM   #5
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Yes, that's why.
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