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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 \(2^{n+1}.\)
Apparently this can be solved with some calculus, but if possible, could you please stick to precalculus (or even algebra 2) terms? That is the class that this question is from. Thanks!
zmudz
 one year ago
Let \(n\) be a positive integer. Show that the smallest integer greater than \((\sqrt{3} + 1)^{2n}\) is divisible by \(2^{n+1}.\) Apparently this can be solved with some calculus, but if possible, could you please stick to precalculus (or even algebra 2) terms? That is the class that this question is from. Thanks!

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Loser66
 one year ago
Best ResponseYou've already chosen the best response.1\((\sqrt 3 +1)^{2n}= ((\sqrt 3 +1)^2)^n = (4+2\sqrt 3)^n = 2^n (2 + \sqrt 3)^n\) Consider the ceiling function of \(2+\sqrt 3\), it is \(\lceil (2+\sqrt 3)\rceil\) =4 hence the smallest integer greater than \(2^n(2+\sqrt 3)^n \) is \(2^n 4^n = 2^n 2^{2n}\), which is divided by \(2^{n+1}= 2*2^n\)
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