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I am trying to prove, by induction, that $2^n > n^2 - 7, \forall n \in \mathbb{N}$.

I am stuck in the inductive step: $n = n+1$.

$$\begin{align*} 2^{n+1} &= 2 \cdot 2^n \\ &> 2 \cdot (n^2 - 7) \tag{By I.H.} \\ &= 2 \cdot (n^2 + 2n + 1 - 10) \\ &= 2(n+1)^2 - 20 \\ &> (n+1)^2 - 17 \end{align*} $$

Which is not what I want, $(n+1)^2 - 7$. I'm not seeing how to reduce the constant to 7, any hints appreciated.

  • 2
    The title says $\gt$ but you are proving $\lt$ ? – 2017-02-19
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    $2(n^2-7)\ge(n+1)^2-7$ for $n > 3$, and you can solve it by hand for $n \le 3$. – 2017-02-19
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    A stronger inequality is $2^n>n^2-2$ (for all $n\in\mathbb{N}). – 2017-02-19
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    Why does $n^2-7 =n^2+2n+1-10$? – 2017-02-19

1 Answers 1

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First off, $+1$ for an excellent question.

Second, you have switched signs ($>$ to $<$). You are given that $2^n > n^2-7$,and have to prove it for $n+1$.

To do this, note that: $2^{n+1} = 2 \cdot 2^n > 2(n^2-7) > 2n^2 - 14$.

Note that $2n^2 - 14 - ((n+1)^2 - 7) = (n+2)(n-4) > 0$ if $n \geq 5$.

Hence, it follows that $2^{n+1} > 2n^2 - 14 > (n+1)^2 - 7$ for $n \geq 5$. You can check the rest manually (I leave it to you).

Key point : Don't worry if the RHS doesn't immediately suggest a finish to the argument. In this case, I took the difference between the RHS I got, and the RHS I wanted, and the nature of the difference allowed me to deduce the identity for all but a very small set of small numbers, which was easy to do manually.

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    Do keep this trick in mind! – 2017-02-19
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    I certainly will. Thank you. – 2017-02-19
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    You are welcome @mnk888 – 2017-02-19