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I have a problem that I have to solve using mathematical induction but I'm stuck from a part. The problem is:

Proof that $\large n<2^n$ is true for $\large n \in \mathbb{N}\ $

So, I did that steps:

1) $n = 1$, $\qquad$ $1<2^1$ $\rightarrow$ $1 < 2$ $\qquad$ THIS IS TRUE

2) $n = k$, $\qquad$ $k < 2^k$

3) $n = k+1$, $\qquad$ $k+1 < 2^{k+1}\ $

$\qquad$$\qquad$$\qquad$$\quad$ $k < 2^k$

$\qquad$$\qquad$$\qquad$$\quad$ $2 * k < 2^k * 2$ $\qquad$ I multiplied by $2^1$ in both sides to get

$\qquad$$\qquad$$\qquad$$\quad$ $2k < 2^{k+1}$

I'm stuck in this part, what should I do next?? Please be specific

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  • $\begingroup$ If $k>1$ then $k+1<k+k=2k$, so ... $\endgroup$ – DiegoMath Sep 7 '14 at 6:38
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When you have $k<2^k$ (2) then $k+\color{red}1<2^k+\color{red}1$. But $$\color{red}1< \color{green}{2^k},~~\forall k\in\mathbb N.$$ so you get $$k+\color{red}1<2^k+\color{red}1<2^k+\color{green}{2^k}.$$

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For $n=1$ we have $$1 < 2.$$ From $n=k$ to $n=k+1$ note that $k < 2^{k}$ is equivalent to $k+1 \leq 2^{k},$ so that $$k+1 \leq 2^{k} < 2^{k+1}$$ follows.

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I think you should have done it like this

By assumption:

          k+1<2^k


         2(k+1)<2*2^k            * by 2 both sides

since 2*2^k>2(k+1)

and 2(k+1)>k+1

  for example if n>10


  therefore     n>5

Therefore k+1<2*2^k

         =2^(k+1)

LHS < RHS

Then statement n<2^n is valid for all Natural Numbers


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