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Thread: Size!

  1. November 10th 2009, 12:33 PM #1
    Drexel28
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    Size!

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

    Extra: Can a generlization be made about the bases?
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  2. November 11th 2009, 03:16 AM #2
    Deadstar
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    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 99^n + 100^n > 101^n to 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.
    Last edited by Deadstar; November 11th 2009 at 03:33 AM.
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  3. November 11th 2009, 04:25 AM #3
    Deadstar
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    Umm... To further my ramblings...

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

    i.e. that 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
    => 1 + 0.99^n < 1.01^n.

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

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

    Pretty sure this is 49 but I don't know how to provide a proper proof!
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  4. November 11th 2009, 08:14 AM #4
    Drexel28
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    Quote Originally Posted by Deadstar View Post
    Umm... To further my ramblings...

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

    i.e. that 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
    => 1 + 0.99^n < 1.01^n.

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

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

    Pretty sure this is 49 but I don't know how to provide a proper proof!
    Spoiler:
    It is indeed true if n\ge 49! Now try to find a nice proof for it.
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  5. November 16th 2009, 03:40 PM #5
    Drexel28
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    Partial solution

    Spoiler:
    The above inequality is true when n\ge 49

    Proof: This is equivalent to showing that \frac{101^n=99^n}{100^n}>1. But 101^n-99^n=\left(100+1\right)^n-\left(100-1\right)^n=\sum_{\ell=1}^{n} {n \choose \ell}100^{n-\ell} -\sum_{\ell=0}^{n} {n\choose \ell}100^{n-\ell}(-1)^{\ell}=\sum_{n=0}^{n}{n\choose \ell}100^{n-\ell}\left(1-\left(-1\right)^{\ell}\right)\color{red}\star. Now clearly 1-(-1)^{\ell}=\begin{cases} 0 & \mbox{if}\quad \ell\text{ is even} \\ 1 & \mbox{if}\quad \ell\text{ is odd} \end{cases}. So \color{red}\star reduces to 2\sum_{\ell=0}^{\frac{n-1}{2}}{n\choose {2\ell-1}}100^{n-2\ell+1}=2\left(100^{n-1}n+\frac{n(n-1)(n-2)100^{n-3}}{ 6}+\cdots\right). Therefore \frac{101^n-99^n}{100^n}=2\left(\frac{n}{100}+\frac{n(n-1)(n-2)}{100^3\cdot 6}+\cdots\right). If n\ge50 this is trivial since \frac{101^n-99^n}{100^n}>2\cdot\frac{n}{100}>1. For the case where n=49 merely note that 2\cdot\left(\frac{49}{100}+\frac{49\cdot48\cdot47} {100^3\cdot 6}+\cdots\right)>2\cdot\left(\frac{49}{100}+\frac{ 49\cdot48\cdot47}{100^3\cdot 6}\right)>2\left(\frac{49}{100}+\frac{1}{100}\righ t)>1
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