# Proving that $2^n+1\leq 3^n$ by induction [closed]

I need to prove the following using mathematical induction:

$$2^n+1\leq 3^n\qquad\forall n\in\Bbb{Z^+}$$

Been working on this problem for a while and cannot figure it out. Any guidance or help would be appreciated on how to start or complete the inductive step, the part that is tripping me up.

## closed as off-topic by user223391, user228113, user147263, zarathustra, JonMark PerryDec 27 '15 at 8:50

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• using induction i got 4^k+1<=9^k – barbarian Apr 16 '15 at 20:04
• @avid19 The hope is that your comment would be true, but more often than not it is not really true. Usually if there is a question like this...I prefer to either give a partial answer (as I have done here), downvote, downvote and vote to close, flag, etc. Think to yourself like the Count of Monte Cristo: "Do you worst, for I will do mine!" – Daniel W. Farlow Apr 16 '15 at 20:32

For $n=1$, $2^1+1=3=3^1$. It is proved for $n=1$
Assume it is true for $n$.
$2^{n+1}+1=2(2^{n}+1)-1\leq 2\cdot3^n-1\leq 2\cdot 3^n<3^{n+1}$
The inequality is trivial for $n=1$. Suppose it holds for $n=p$. Then $$2^{p+1}+1 = 2\cdot 2^p+1 =2^p+2^p+1$$ which by assumption $$\leq 2^p+3^p \leq 3^p+3^p=2\cdot 3^p\leq 3\cdot 3^p=3^{p+1}$$ so the inequality holds for $n=p+1$. The induction axiom then tells that the inequality is true for any natural number.