Consider the set \(S_n\) of all the \(2^n\) numbers of the type \(2\pm\sqrt{2\pm\sqrt{2\pm\dots}}\), where the number \(2\) appears \(n+1\) times.

(a) Show that all members of \(S_n\) are real.

(b) Find the product \(P_n\) of all elements of \(S_n\).

Source: Austria 1989

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TopNewest(a)Assume, up to a number n, that all values are positive and real.These values are bounded by 4, since the infinite radical \(\sqrt{2+\sqrt{2+...}}=2\).So the n+1st number is bigger than \(2-\sqrt{2+\sqrt{2+...}}=0\).Thus, by induction, every such number is real and positive (the base case is obvious)

(b)I will show that the product is the same for every n.Let \(\alpha_n\) be a number \(\sqrt{2\pm\sqrt{2\pm...}}\).That means that\( P_n=\displaystyle\prod_{}{}(2+\alpha_n)(2-\alpha_n)=\prod 4-\alpha_n^2=\prod (2+\alpha_{n-1})(2+\alpha_{n-1})=P_{n-1}\).Since \(P_1=2\), they are all 2. – Bogdan Simeonov · 2 years, 3 months ago

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– Cody Johnson · 2 years, 3 months ago

Bravo !Log in to reply

– Bogdan Simeonov · 2 years, 3 months ago

Thanks :D .By the way when is the next Proofathon?Log in to reply

– Cody Johnson · 2 years, 3 months ago

See the schedule on the website: http://proofathon.org/ongoing_contest.phpLog in to reply