### Proove that $2n-3\leq 2^{n-2}$ for all $n\geq 5$

The following is the proof asked in https://goo.gl/h5uyhK

Prove that $$2n-3\leq 2^{n-2}\text{ for all } n\geq 5$$

Solution (by Mathematical Induction)

1. Base case: If $n=5$, the left side is 7 and the right side is $2^{5-2}=2^3=8$. So, the inequality holds when $n=5$.

2. Inductive hypothesis: Suppose the inequality holds for all values of $n$ up to some $k\geq5$, $P(k):2k-3\leq 2^{k-2}$.
3. Inductive step: Let $n=k+1$. Now we have to prove that the relation also holds for $k+1$ by using the induction hypothesis. This means that we have to prove
$$P(k+1):2(k+1)-3\leq 2^{(k+1)-2}$$
Then our left side is
\begin{align} P(k+1):2(k+1)-3 &= 2k-3+2 \\ &\leq 2^{k-2} +2 \text{, by our inductive hypothesis}\\ &\leq 2^{k-2} + 2^{k-2} \text{ since } 2\leq 2^{k-2}, \forall k\geq 5 \\ &\leq 2\cdot 2^{k-2} \\ &\leq 2^{k-2+1} \\ &=2^{k-1} \end{align}
Thus
$$2(k+1)-3\leq 2^{k-1} \Longrightarrow 2k-3 \leq 2^{k-2}$$
which is our right side. So, the theorem holds for $n=k+1$.
By the principle of mathematical induction, the theorem holds for all $n\geq 5$.

### Comentarios

1. Awesome proof and $\LaTeX$ display!

2. Nice proof. Your proof is helpfull for me. I can understand this explanation too. Thank you very much for your help.

1. You are welcome Dian! I'm glad it helped you.

3. Okay Luis. I'll ask you if I need a help. Thank you so much Luis. I'm not good at mathematical induction :'(

1. Be my guest!

And do not worry, it could be difficult at the beggining, but you can get use to it whenever you practice what you need. Keep going!

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