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  • 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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  1. Loser66
    • one year ago
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    \((\sqrt 3 +1)^{2n}= ((\sqrt 3 +1)^2)^n = (4+2\sqrt 3)^n = 2^n (2 + \sqrt 3)^n\) 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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