Let $\displaystyle n\in\mathbb{N}$. Which is larger $\displaystyle 99^n+100^n$ or $\displaystyle 101^n$? (there may be more than one case)Problem:

Can a generlization be made about the bases?Extra:

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- Nov 10th 2009, 12:33 PMDrexel28Size!
Let $\displaystyle n\in\mathbb{N}$. Which is larger $\displaystyle 99^n+100^n$ or $\displaystyle 101^n$? (there may be more than one case)**Problem:**

Can a generlization be made about the bases?__Extra:__ - Nov 11th 2009, 03:16 AMDeadstar
Ok this hasn't been thought out well but I thought I would write down my initial thoughts.

For an n you have 1 set of (99 + 100) and 1 set of 101 on its own (I can't explian whats going on in my head for this part!).

The difference between the two is 98 so my guess is the change from $\displaystyle 99^n + 100^n > 101^n$ to $\displaystyle 99^n + 100^n < 101^n$ would happen at n=49.

Why I cant fully explain but something along the lines of that there's two things n acts on in on the LHS and one on the RHS and the RHS will 'catch' up to the LHS by 2 each time........

Ill think some more about what I'm actually trying to say but for now I think the change happens at n=49. - Nov 11th 2009, 04:25 AMDeadstar
Umm... To further my ramblings...

Want to check if $\displaystyle 99^n + 100^n - 101^n < 0$.

i.e. that $\displaystyle 99^n + 100^n - 101^n = 100^n(0.99^n + 1^n - 1.01^n) = 100^n(0.99^n + 1 - 1.01^n) < 0$

$\displaystyle => 1 + 0.99^n < 1.01^n$.

Clearly $\displaystyle 1 + 0.99^n > 1$ and $\displaystyle 1.01^n > 1$, $\displaystyle 1 + 0.99^n$ is converging towards 1 while $\displaystyle 1.01^n$ is monotonically increasing.

Hence at some point $\displaystyle 1.01^n$ will exceed $\displaystyle 1 + 0.99^n$ for some value of $\displaystyle n$. Just remains to find this $\displaystyle n$ as up until this point $\displaystyle 99^n + 100^n > 101^n$.

Pretty sure this is 49 but I don't know how to provide a proper proof! - Nov 11th 2009, 08:14 AMDrexel28
- Nov 16th 2009, 03:40 PMDrexel28
Partial solution

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