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I'm reading through an example proof for proving $3^n > 2n^2 + 3n$ for $n\ge4 $ I just don't understand how the example goes from step 3 to 4. Everything before and after that step makes sense to me.

  1. $3 * 3^n > 6n^2 + 9n$

  2. $3^{n+1} > 6n^2 + 9n$

  3. $\qquad\;= 2(n^2 + 2n + n) + (4n^2 + 3n) $

  4. $\qquad\;> 2(n^2 + 2n + 1) + (3n + 3)$

  5. $\qquad\;=2(n+ 1)^2 + 3(n + 1)$

How do I go from step 3 to step 4?

2 Answers 2

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It seems that this is part of an induction proof, to prove $3^n > 2n^2 + 3n$. I am just mentioning it to put these steps into a greater context.

Notice that from step 3 to step 4 we go from an equality to an inequality. The term in the RHS of the equation is transformed (reduced) to make it strictly smaller than the term in the RHS of step 3. The transformation has the final goal of producing $2(n+ 1)^2 + 3(n + 1)$. Do you see why?

Now this step is correct since: $$2(n^2 + 2n + n)>2(n^2 + 2n + 1) \qquad \text{since}\;\; n>1$$ and $$4(n^2 + 3n) = 4n^2 + 12n > 3n +3 \qquad \text{since}\;\; 4n^2 > 3n \;\;\text{and}\;\;12n>3$$

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    Ah ok, it's just the formatting that makes it confusing. Writing $2(n^2 + 2n + n) + (4n^2 + 3n) > 2(n^2 + 2n + 1) + (3n + 3)$ would be a little clearer. Are there any special rules for going from an equality to an inequality? – 2017-02-20
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    @ZeroPhase I am wondering why you were not confused by the transition from step 2 to step 3. This seems more arbitrary to me (if you do not have a sense of the general direction of the proof). Did you just miss the > sign on step 4? – 2017-02-20
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    Not sure what you mean by "special rules". You just try to create something smaller (or bigger) and try to do it in steps that are obvious to your audience. – 2017-02-20
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    I could just see how step 3 could be arrived at by rearranging the terms. Noticed I actually had a typo in step 3. By special rules I mean if there was anything similar to flipping the comparison sign for going from equals to a comparison. – 2017-02-20
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    @ZeroPhase There's no such thing as "going from an equality to an inequality". It's just that the thing on line $3$ is bigger than the thing on line $4$. Line $4$ doesn't "remember" that line $3$ was part of an equality. – 2017-03-08
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You have

$2(n^2 + 2n + n)>2(n^2 + 2n + 1)$ since $n>3$

and

$4(n^2 + 3n)> 4n^2 + 12n > 3n^2+3n > 3n +3$

again because $n>3$.