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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 \(\lceil (\sqrt{3}+1)^{2n} \rceil = (\sqrt{3}+1)^{2n} + (\sqrt{3}1)^{2n}.\)
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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amilapsn
 one year ago
Best ResponseYou've already chosen the best response.3you can use mathematical induction with binomial expansion.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Id probably say make a recurrence relation

zzr0ck3r
 one year ago
Best ResponseYou've already chosen the best response.1A number can't be equal to itself plus a nonzero number...

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Isn't that\[\lceil (\sqrt{3}+1)^{2n} \rceil = (\sqrt{3}+1)^{n} + (\sqrt{3}1)^{n}.\]

amilapsn
 one year ago
Best ResponseYou've already chosen the best response.3\[\exists ! 0.abcd.... ~s.t.\] \[\lceil{(\sqrt{3}+1)^{2n}}\rceil=(\sqrt{3}+1)^{2n}+0.abcd.....\] \[But~ 0<(\sqrt{3}1)^{2n}< 1\forall n\in \mathbb{Z}^+\] By binomial expansion: \[\small (\sqrt{3}+1)^{2n}=^{2n}C_0 (\sqrt{3})^{2n}+^{2n}C_1 (\sqrt{3})^{2n1}+\ldots+^{2n}C_{2n1} (\sqrt{3})^{1}+^{2n}C_{2n} 1 ~~(2n+1~Terms)\] \[\small (\sqrt{3}1)^{2n}=^{2n}C_0 (\sqrt{3})^{2n}^{2n}C_1 (\sqrt{3})^{2n1}+\ldots^{2n}C_{2n1} (\sqrt{3})^{1}+^{2n}C_{2n} 1 ~~(2n+1~Terms)\] \[\small \therefore (\sqrt{3}+1)^{2n}+(\sqrt{3}1)^{2n}=2 \left[^{2n}C_0 (\sqrt{3})^{2n}+^{2n}C_{2} (\sqrt{3})^{2n2}+\ldots+^{2n}C_{2n2} (\sqrt{3})^{2}+^{2n}C_{2n} 1\right] \] 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.. \[2^{n+1}\lceil{(\sqrt{3}+1)^{2n}}\rceil\]

zzr0ck3r
 one year ago
Best ResponseYou've already chosen the best response.1suppose \(a=a+b\) where \(b\ne 0\). Then by cancelation \(b=0\) a contradiction. :)

amilapsn
 one year ago
Best ResponseYou've already chosen the best response.3@zzr0ck3r you've forgotten the ceiling... :) for example \(\lceil 2.5\rceil =3\)

zzr0ck3r
 one year ago
Best ResponseYou've already chosen the best response.1Ahh, sorry on my phone and I thought it was just \([\). lol. I was going nuts...
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