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I am unsure how to prove the following and any help would be appreciated.

Prove that $2^{3n} > 3 + 4n$ for all $n$ is greater than or equal to $1$.

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closed as off-topic by John Omielan, Shailesh, Ak19, Jendrik Stelzner, zoli Jun 13 at 15:50

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You haven't shown any work, or indicated where you are stuck. So I will just give you...

Hint: To go from $n$ to $n+1$: $2^{3(n+1)}=2^{3n+3}=8\cdot 2^{3n}$

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The base case $n=1$ holds, because $8>7$. Now we assume the statement holds for $n$, and want to prove it still holds for $n+1$. $$\begin{array}{rcl}2^{3(n+1)}&=&2^{3n+3}\\&=&8\cdot 2^{3n}\\&>&8\cdot (3+4n)\quad(\text{by induction hypothesis})\\&>&24+4n\\&=&20+4(n+1)\\&>&3+4(n+1)\end{array}$$ Does this help?

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  • $\begingroup$ yes so in conclusion, because 28n + 17 greater than or equal to 0, which is true. Therefore 2^(3n+3) > 3+4(n+1). By gen pmi, 2^(3n) > 3+4n for all n greater than or equal to 1. $\endgroup$ – kengriffinuchi21 Jun 12 at 13:31
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Base case: For $n=1$ we get $8>7$ which is true. Now we assume that for $n=k$ is hold $$2^{3k}>3+4k$$ and we have to prove, that $$2^{3k+3}>4+4(k+1)$$ Multiplying the second inequality above with $2^3$ we get $$3^{3k+3}>24+32k$$ now we must prove that $$24+32k>3+4(k+1)$$ Can you finish?

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  • $\begingroup$ yes so in conclusion, because 28n + 17 greater than or equal to 0, which is true. Therefore 2^(3n+3) > 3+4(n+1). By gen pmi, 2^(3n) > 3+4n for all n greater than or equal to 1. $\endgroup$ – kengriffinuchi21 Jun 12 at 13:48
  • $\begingroup$ is my conclusion correct? $\endgroup$ – kengriffinuchi21 Jun 12 at 13:48
  • $\begingroup$ Yes, this is correct! $\endgroup$ – Dr. Sonnhard Graubner Jun 12 at 14:27
  • $\begingroup$ Thank you, this is supposed to be gen PMI right? not PCI? $\endgroup$ – kengriffinuchi21 Jun 12 at 15:43
  • $\begingroup$ What does PMI mean? $\endgroup$ – Dr. Sonnhard Graubner Jun 12 at 15:45

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