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If $n$ be a positive integer greater than $1$, then prove that $$2^n>1+n\sqrt{2^{n-1}}$$

I found this problem under $AM \ge GM$ chapter. Help me to solve this problem using $AM \ge GM$. Thanks in advance.

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1 Answer 1

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Hint:$1+2+2^2+\dots 2^{n-1}=2^n-1$

Edit:

Solution:

Applying A.M.-G.M. on $\displaystyle \sum_{k=0}^{n-1}2^{k}$ we have $2^n-1=\displaystyle \sum_{k=0}^{n-1}2^{k}\ge n(2^{(\sum_{k=0}^{n-1}k)})^{\frac{1}{n}}=n2^{\frac{n-1}{2}}$

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    $\begingroup$ Again you beat me :) (+1) $\endgroup$ Jun 19, 2013 at 10:45
  • $\begingroup$ [email protected]. Thank by the way... $\endgroup$ Jun 19, 2013 at 10:46
  • $\begingroup$ It almost hurts for its simplicity. +1 $\endgroup$
    – DonAntonio
    Jun 19, 2013 at 10:49
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    $\begingroup$ Thank you for helping me. This is the best answer for this question. $\endgroup$
    – A.D
    Jun 19, 2013 at 10:56
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    $\begingroup$ @A.D I think you are correct, considering it's the only answer. (tongue in cheek) $\endgroup$
    – Thomas
    Jun 19, 2013 at 12:21

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