# Prove the following with Generalized PMI [closed]

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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## 3 Answers

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

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

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?

• 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. – kengriffinuchi21 Jun 12 at 13:31

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?

• 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. – kengriffinuchi21 Jun 12 at 13:48
• is my conclusion correct? – kengriffinuchi21 Jun 12 at 13:48
• Yes, this is correct! – Dr. Sonnhard Graubner Jun 12 at 14:27
• Thank you, this is supposed to be gen PMI right? not PCI? – kengriffinuchi21 Jun 12 at 15:43
• What does PMI mean? – Dr. Sonnhard Graubner Jun 12 at 15:45