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How can I show that:

$ a^n-1 \geq n\left(a^{\frac{n+1}{2}}-a^\frac{n-1}{2}\right)$

$ \sum_{k=0}^{n-1} a^k \geq na^\frac{n-1}{2}$

$ a>1, n\in\mathbb{N} $

without studying the function

$ f(x)=x^n-1 - n\left(x^{\frac{n+1}{2}}-x^\frac{n-1}{2}\right)$?

1 Answers 1

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$a^n-1=(a-1)(a^{n-1}+a^{n-2}+..+a+1)$

By AM-GM

$a^{n-1}+a^{n-2}+..+a+1 > na^{\frac{n-1}{2}}$

Thus

$a^n-1 >(a-1)na^{\frac{n-1}{2}}$

  • 0
    Thank you for your answer!2012-08-28