Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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.

share|improve this question

1 Answer 1

up vote 19 down vote accepted

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 $2^n-1=\displaystyle \sum_{k=0}^{n-1}2^{k}\ge n(2^{(\sum_{k=0}^{n-1}k)})^{\frac{1}{n}}=n2^{\frac{n-1}{2}}$

share|improve this answer
2  
Again you beat me :) (+1) –  O.L. Jun 19 '13 at 10:45
    
Really???@O.L. Thank by the way... –  Abhra Abir Kundu Jun 19 '13 at 10:46
    
It almost hurts for its simplicity. +1 –  DonAntonio Jun 19 '13 at 10:49
    
Thanks @DonAntonio –  Abhra Abir Kundu Jun 19 '13 at 10:50
2  
Thank you for helping me. This is the best answer for this question. –  A.D Jun 19 '13 at 10:56

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.