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zmudz

  • one year ago

Let \(n\) be a positive integer. Show that the smallest integer greater than \((\sqrt{3} + 1)^{2n}\)is divisible by \(2^{n+1}.\) Hint: Prove that \(\lceil (\sqrt{3}+1)^{2n} \rceil = (\sqrt{3}+1)^{2n} + (\sqrt{3}-1)^{2n}.\)

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  1. amilapsn
    • one year ago
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    you can use mathematical induction with binomial expansion.

  2. anonymous
    • one year ago
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    Id probably say make a recurrence relation

  3. zzr0ck3r
    • one year ago
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    A number can't be equal to itself plus a nonzero number...

  4. anonymous
    • one year ago
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    Isn't that\[\lceil (\sqrt{3}+1)^{2n} \rceil = (\sqrt{3}+1)^{n} + (\sqrt{3}-1)^{n}.\]

  5. amilapsn
    • one year ago
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    \[\exists ! 0.abcd.... ~s.t.\] \[\lceil{(\sqrt{3}+1)^{2n}}\rceil=(\sqrt{3}+1)^{2n}+0.abcd.....\] \[But~ 0<(\sqrt{3}-1)^{2n}< 1\forall n\in \mathbb{Z}^+\] By binomial expansion: \[\small (\sqrt{3}+1)^{2n}=^{2n}C_0 (\sqrt{3})^{2n}+^{2n}C_1 (\sqrt{3})^{2n-1}+\ldots+^{2n}C_{2n-1} (\sqrt{3})^{1}+^{2n}C_{2n} 1 ~~(2n+1~Terms)\] \[\small (\sqrt{3}-1)^{2n}=^{2n}C_0 (\sqrt{3})^{2n}-^{2n}C_1 (\sqrt{3})^{2n-1}+\ldots-^{2n}C_{2n-1} (\sqrt{3})^{1}+^{2n}C_{2n} 1 ~~(2n+1~Terms)\] \[\small \therefore (\sqrt{3}+1)^{2n}+(\sqrt{3}-1)^{2n}=2 \left[^{2n}C_0 (\sqrt{3})^{2n}+^{2n}C_{2} (\sqrt{3})^{2n-2}+\ldots+^{2n}C_{2n-2} (\sqrt{3})^{2}+^{2n}C_{2n} 1\right] \] Since all the terms have even exponents of \(\sqrt{3}\) this will be an integer... \[\therefore \lceil{(\sqrt{3}+1)^{2n}}\rceil=(\sqrt{3}+1)^{2n}+(\sqrt{3}-1)^{2n}\] By mathematical induction we can show that.. \[2^{n+1}|\lceil{(\sqrt{3}+1)^{2n}}\rceil\]

  6. zzr0ck3r
    • one year ago
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    suppose \(a=a+b\) where \(b\ne 0\). Then by cancelation \(b=0\) a contradiction. :)

  7. amilapsn
    • one year ago
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    @zzr0ck3r you've forgotten the ceiling... :) for example \(\lceil 2.5\rceil =3\)

  8. zzr0ck3r
    • one year ago
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    Ahh, sorry on my phone and I thought it was just \([\). lol. I was going nuts...

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